Math worksheet focusing on perpendicular bisectors in triangles, including diagrams and measurement exercises.

Let’s solve each problem step by step.

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Problem 1: Find each length.

We are given a triangle with a perpendicular bisector from point D to side AC, and we’re told that DB is the perpendicular bisector of AC. That means:

- It cuts AC into two equal parts → so AB = BC
- Also, since it’s a perpendicular bisector in a triangle, if D is on the bisector, then DA = DC (because any point on the perpendicular bisector of a segment is equidistant from the endpoints).

Given:
- DA = x + 4
- DC = 3x - 2

Since DA = DC (because D is on the perpendicular bisector of AC):

→ x + 4 = 3x - 2

Solve for x:

Subtract x from both sides:
4 = 2x - 2

Add 2 to both sides:
6 = 2x

Divide by 2:
x = 3

Now find the lengths:

DA = x + 4 = 3 + 4 = 7
DC = 3x - 2 = 9 - 2 = 7 ✔ (matches)

Also, AB = BC because DB bisects AC.

AB = 5y - 1
BC = 3y + 5

Set them equal:

5y - 1 = 3y + 5

Subtract 3y:
2y - 1 = 5

Add 1:
2y = 6

Divide by 2:
y = 3

Then:

AB = 5(3) - 1 = 15 - 1 = 14
BC = 3(3) + 5 = 9 + 5 = 14 ✔

So answers for Problem 1:

a. DA = 7
b. DC = 7
c. AB = 14
d. BC = 14

Wait — looking back at the image layout, it says “Find each length” and lists:

a. ___
b. ___
c. ___
d. ___

And labels next to diagram: DA = x+4, DC=3x-2, AB=5y-1, BC=3y+5

So likely:

a. DA
b. DC
c. AB
d. BC

But let’s check the numbering in the original worksheet — actually, looking again, the first diagram has points A, B, C, D, and segments labeled. The questions are probably asking for those four lengths.

So final for #1:

a. DA = 7
b. DC = 7
c. AB = 14
d. BC = 14

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Problem 2: Find each length.

Diagram shows triangle JKL, with M on JL, and KM is drawn. Given:

JM = 8
ML = 8 → so M is midpoint of JL → JM = ML = 8 → JL = 16

Also, angle at M is marked as right angle → so KM ⊥ JL → so KM is perpendicular bisector? Not necessarily unless KJ = KL.

Wait — look at labels:

KJ = 3z + 1
KL = 5z - 7

If KM is perpendicular to JL and M is midpoint, then if KJ = KL, triangle is isosceles, but here they give expressions — so maybe we assume KM is perpendicular bisector? But only if KJ = KL.

Actually, re-read: In problem 2, it just says “Find each length.” and gives diagram with:

- JM = 8, ML = 8 → so JL = 16
- KJ = 3z + 1
- KL = 5z - 7
- And KM is drawn, with right angle at M → so KM ⊥ JL

But unless told otherwise, we can’t assume KJ = KL. Wait — perhaps in this context, since M is midpoint and KM is perpendicular, then K is on the perpendicular bisector → so KJ should equal KL.

Yes! If a point lies on the perpendicular bisector of a segment, it is equidistant from the endpoints.

Here, M is midpoint of JL, and KM ⊥ JL → so line KM is the perpendicular bisector of JL → therefore, any point on KM (including K) is equidistant from J and L → so KJ = KL.

Therefore:

3z + 1 = 5z - 7

Solve:

Subtract 3z:
1 = 2z - 7

Add 7:
8 = 2z

Divide by 2:
z = 4

Then:

KJ = 3(4) + 1 = 12 + 1 = 13
KL = 5(4) - 7 = 20 - 7 = 13 ✔

Also, JM = 8, ML = 8 → already given.

The question says “Find each length.” and lists:

a. ___
b. ___

Looking at diagram, likely:

a. KJ or KL
b. JL? Or maybe KM? But KM not given expression.

Wait — in the diagram, only KJ and KL have variables. JM and ML are given as 8.

Perhaps the blanks are for KJ and KL? Or maybe JL?

But JL = JM + ML = 8 + 8 = 16 — no variable.

Probably the unknowns are KJ and KL, which we found as 13 each.

But let’s see how the worksheet is structured — in problem 1, there were 4 blanks for 4 expressions. Here, only 2 blanks.

Likely:

a. KJ = 13
b. KL = 13

Or perhaps one is JL? But JL is 16, straightforward.

Wait — looking back at user's image description, problem 2 says “Find each length.” and has two blanks: a. ___ b. ___

And in diagram, KJ and KL are labeled with expressions, while JM and ML are numbers.

So most likely, a and b refer to KJ and KL.

So:

a. KJ = 13
b. KL = 13

But to be precise, let’s note that JL = 16, but it’s not asked since no blank for it.

Alternatively, maybe a is JL and b is something else? Unlikely.

I think safe to say:

For problem 2:

a. KJ = 13
b. KL = 13

But let’s double-check — is there another length? KM is not given, so probably not.

Perhaps the blanks correspond to the sides with variables.

Yes.

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Problems 3–6: Use the diagram below. XY is the perpendicular bisector of WZ. Find the indicated measure.

Diagram: Points W, X, Y, Z. XY is perpendicular bisector of WZ → so it cuts WZ at its midpoint, say M, and XY ⊥ WZ.

Also, angles and sides labeled:

Angle at X: ∠WXY = 5x + 6
Angle at Y: ∠ZYX = 7x - 2
Side WX = 10x - 14
Side ZY = 8x + 2

Since XY is perpendicular bisector of WZ, then:

- Any point on XY is equidistant from W and Z.

So, for example, point X is on XY → so XW = XZ
Point Y is on XY → so YW = YZ

Also, since it’s perpendicular bisector, triangles WXM and ZXM are congruent, etc.

But more directly:

Because X is on perpendicular bisector of WZ → XW = XZ
Similarly, Y is on it → YW = YZ

But in the diagram, we have WX and ZY given — wait, WX is same as XW, ZY is same as YZ.

So:

XW = XZ → but we don’t have XZ labeled. We have WX = 10x - 14, and ZY = 8x + 2.

But ZY is YZ, which equals YW, not necessarily related to XW.

Wait — perhaps we need to use angles.

Note that since XY is perpendicular bisector, and assuming it intersects WZ at M, then WM = MZ, and angles at M are 90 degrees.

Also, triangles WXY and ZXY might be congruent? Not necessarily.

Another approach: since XY is perpendicular bisector, then reflection over XY maps W to Z.

Therefore, distances and angles should match accordingly.

Specifically, angle at X: ∠WXY should equal angle at X for Z, but since it's symmetric, actually ∠WXY = ∠ZXY? Not exactly.

Let me denote:

Let M be the intersection of XY and WZ. Since XY is perp bisector, M is midpoint, and XY ⊥ WZ.

Now, consider triangles WXM and ZXM:

- WM = ZM (bisected)
- XM common
- Angle at M is 90 for both → so by SAS, triangles WXM ≅ ZXM → so WX = ZX, and angles correspond.

Similarly, for Y: triangles WYM and ZYM ≅ → so WY = ZY.

Ah! So:

WX = ZX
WY = ZY

But in the diagram, we are given:

WX = 10x - 14
ZY = 8x + 2

But ZY is same as YZ, which equals WY.

So WY = ZY = 8x + 2

But we don't have direct relation between WX and WY yet.

However, we also have angles:

∠WXY = 5x + 6
∠ZYX = 7x - 2

Note that ∠ZYX is the same as ∠WYX? No.

Actually, since the figure is symmetric over XY, then angle between WX and XY should equal angle between ZX and XY.

That is, ∠WXY = ∠ZXY

Similarly, ∠WYX = ∠ZYX

Is that correct?

Yes, because reflection over XY swaps W and Z, so angles at X and Y should be symmetric.

Therefore:

∠WXY = ∠ZXY
But ∠ZXY is not labeled; instead, we have ∠ZYX.

∠ZYX is at Y, between ZY and YX.

By symmetry, ∠WYX should equal ∠ZYX.

But we are given ∠WXY and ∠ZYX.

Are these related?

Not directly, unless we consider triangle WXY and ZXY.

Another idea: since WX = ZX and WY = ZY, and XY common, then triangles WXY and ZXY are congruent by SSS.

Therefore, corresponding angles are equal.

So, angle at X in triangle WXY equals angle at X in triangle ZXY → so ∠WXY = ∠ZXY

Similarly, angle at Y: ∠WYX = ∠ZYX

But we are given ∠WXY = 5x + 6 and ∠ZYX = 7x - 2

Note that ∠ZYX is the same as ∠WYX by symmetry? Yes, because reflection swaps W and Z, so angle between WY and YX equals angle between ZY and YX.

So ∠WYX = ∠ZYX = 7x - 2

In triangle WXY, we have angles at X and Y: ∠WXY = 5x + 6, ∠WYX = 7x - 2

And angle at W is unknown, but sum of angles in triangle is 180.

But we don't know angle at W.

However, we also have side lengths: WX = 10x - 14, WY = ? , XY = ?

But we know that in triangle WXY, sides opposite angles.

Perhaps better to use the fact that since triangles WXY and ZXY are congruent, and we have expressions, but we need another equation.

Notice that we have two expressions involving x: one for angle at X, one for angle at Y, and also for sides.

But the key is that for point X, since it's on perpendicular bisector, WX = ZX, but we don't have ZX.

Similarly for Y, WY = ZY = 8x + 2

But we have WX = 10x - 14

No direct equality between WX and WY.

Unless... perhaps in the diagram, there is more.

Another thought: perhaps the angles given are such that we can set up an equation using the property that the perpendicular bisector creates isosceles triangles or something.

Let's list what we know:

From perpendicular bisector XY of WZ:

1. WX = ZX (since X on perp bisector)
2. WY = ZY (since Y on perp bisector)
3. Angles: by symmetry, ∠WXY = ∠ZXY, and ∠WYX = ∠ZYX

Given:
∠WXY = 5x + 6
∠ZYX = 7x - 2

But ∠ZYX = ∠WYX, as established.

In triangle WXY, the three angles are:

- At X: ∠WXY = 5x + 6
- At Y: ∠WYX = 7x - 2
- At W: let's call it θ

Sum: (5x+6) + (7x-2) + θ = 180
12x + 4 + θ = 180
θ = 176 - 12x

But we don't know θ, so not helpful yet.

Perhaps we can use the side lengths.

We have WX = 10x - 14
WY = ZY = 8x + 2 (since WY = ZY)

In triangle WXY, we have sides WX, WY, and XY.

But we don't have XY.

However, if we assume that the triangle is isosceles or something, but not necessarily.

Another idea: perhaps the angles at X and Y are related because of the perpendicularity.

Recall that XY is perpendicular to WZ, so at point M (intersection), angle is 90 degrees.

But M is on WZ and on XY.

In triangle WXM, angle at M is 90 degrees, so angles at W and X add to 90.

Similarly for other triangles.

Let me define M as the foot of the perpendicular from X and Y to WZ, but since XY is the line, and it's straight, M is the same point for both if X and Y are on the same line, which they are.

Assume that XY intersects WZ at M, and M is between W and Z, and XY ⊥ WZ at M.

Then, in triangle WXM, angle at M is 90°, so:

∠WXM + ∠XWM = 90°

Similarly, in triangle ZXM, same thing.

But ∠WXM is part of ∠WXY.

Actually, ∠WXY is the angle at X in triangle WXY, which is the same as ∠WXM if M is on XY, which it is.

Since XY is a straight line, and M is on it, then ray XM is along XY, so ∠WXY is indeed the angle between WX and XY, which is the same as angle between WX and XM, so in triangle WXM, angle at X is ∠WXM = ∠WXY = 5x + 6

Similarly, in triangle WXM, angle at M is 90°, so angle at W is 90° - (5x + 6)

Similarly, for point Y: in triangle WYM, angle at M is 90°, angle at Y is ∠WYM = ∠WYX = 7x - 2 (since Y is on XY, and M on XY, so ray YM is along YX)

So in triangle WYM, angle at Y is 7x - 2, angle at M is 90°, so angle at W is 90° - (7x - 2) = 92 - 7x

Now, notice that at point W, the total angle ∠XWY is composed of angle from triangle WXM and triangle WYM.

Specifically, angle at W in the big triangle WXY is the sum of angle in WXM and angle in WYM, because M is on WZ, and X and Y are on the other side, so rays WX and WY are on different sides of WM.

Actually, depending on the configuration, but typically, if X and Y are on the same side or opposite.

In standard diagram for perpendicular bisector, often X and Y are on the same line perpendicular to WZ at M, so if X and Y are distinct points on the perpendicular line, then from W, the angle to X and to Y might be adjacent.

To simplify, the angle at W in triangle WXY is the difference or sum of the two small angles.

This is getting messy.

Perhaps a better way: since the figure is symmetric with respect to line XY, then the distance from W to X should equal distance from Z to X, which we have, and similarly for Y.

But we also have that the angle that WX makes with XY should be the same as the angle that ZX makes with XY, which we used.

For the angles given, perhaps we can set the angles at X and Y to be related through the triangle.

Let's calculate the length of WZ or something.

Another idea: perhaps the expressions for the angles are meant to be equal because of some reason, but why would 5x+6 = 7x-2? Let's try that.

Set 5x + 6 = 7x - 2

Then 6 + 2 = 7x - 5x => 8 = 2x => x = 4

Then check if it makes sense.

If x = 4, then:

∠WXY = 5*4 + 6 = 20 + 6 = 26°
∠ZYX = 7*4 - 2 = 28 - 2 = 26°

Oh! Both are 26 degrees.

Is that a coincidence? Probably not.

Why would they be equal? Because of symmetry? But ∠WXY and ∠ZYX are at different vertices.

In the symmetric figure, is there a reason they should be equal?

Let me think: if the figure is symmetric over XY, then the angle that WX makes with XY should equal the angle that ZX makes with XY, which is ∠ZXY, not ∠ZYX.

∠ZYX is at Y.

Perhaps in this specific diagram, with the given labels, when x=4, it works, and also the side lengths make sense.

Let me check the side lengths with x=4.

WX = 10x - 14 = 40 - 14 = 26
ZY = 8x + 2 = 32 + 2 = 34

But earlier we said WY = ZY = 34, and WX = 26

In triangle WXY, sides WX=26, WY=34, and angles at X and Y are both 26 degrees? Sum of angles would be 26+26 + angle at W = 52 + angle W = 180, so angle W = 128 degrees.

Is that possible? Yes, but let's see if it satisfies the perpendicular bisector property.

With x=4, angles are both 26 degrees, which matched when we set them equal.

But why should ∠WXY = ∠ZYX? Is there a geometric reason?

Perhaps because the triangles are similar or something.

Another thought: since XY is perpendicular bisector, and if we consider the whole figure, perhaps triangle WXY and triangle ZYX or something.

Let's calculate the actual values.

Perhaps the intention is that since the setup is symmetric, and the angles are labeled similarly, but in this case, setting them equal gave us x=4, and it worked numerically.

Moreover, in many such problems, they design it so that the angles are equal for simplicity.

So let's proceed with x=4.

Then:

WX = 10*4 - 14 = 40 - 14 = 26
ZY = 8*4 + 2 = 32 + 2 = 34

But ZY = WY, as established, so WY = 34

Now, for the problems:

3. Find YZ

YZ is the same as ZY, which is 34

4. Find WX

WX = 26

5. Find YW

YW is the same as WY, which is 34

6. Find WZ

WZ is the segment being bisected. Since M is midpoint, and we need WM or something.

How to find WZ?

We have points W, M, Z, with M midpoint, and XY perp at M.

In triangle WXM, we have WX = 26, angle at X is 26 degrees, angle at M is 90 degrees.

So in right triangle WXM, angle at X is 26°, hypotenuse is WX = 26

Then, adjacent side to angle X is XM, opposite is WM.

So sin(angle at X) = opposite/hypotenuse = WM / WX

So sin(26°) = WM / 26

Thus WM = 26 * sin(26°)

Similarly, cos(26°) = XM / 26

But we may not need numerical value if we can find exactly.

Note that in triangle WXM, angles are:

At X: 26°
At M: 90°
At W: 180 - 90 - 26 = 64°

Similarly, in triangle WYM, angle at Y is ∠WYX = 26° (since we have x=4, ∠ZYX=26°, and ∠WYX=∠ZYX=26°)

Angle at M is 90°, so angle at W is 64° again? 180-90-26=64°

But at point W, the total angle in triangle WXY is angle between WX and WY.

From above, in triangle WXM, angle at W is 64°, which is the angle between WX and WM.

In triangle WYM, angle at W is 64°, between WY and WM.

Depending on whether X and Y are on the same side of WZ or opposite.

Typically, in such diagrams, if XY is the perpendicular bisector, and X and Y are two points on it, they could be on the same side or opposite sides of WZ.

If they are on the same side, then from W, the rays WX and WY are on the same side of WM, so the angle between them would be the difference of the two angles.

If on opposite sides, sum.

In this case, since both angles at W in the small triangles are 64°, and if X and Y are on the same side of WZ, then angle XWY = |64° - 64°| = 0, impossible.

So likely, X and Y are on opposite sides of WZ.

That makes sense for a perpendicular bisector line; it extends on both sides.

So assume that M is on WZ, and XY is perpendicular at M, with X on one side, Y on the other side of WZ.

Then, from point W, the ray WM is along WZ, and WX is on one side, WY on the other side.

So the angle between WX and WY is the sum of angle between WX and WM and angle between WY and WM.

Which is 64° + 64° = 128°, which matches our earlier calculation for triangle WXY: angles at X and Y are 26° each, so at W is 128°.

Good.

Now, to find WZ.

WZ = WM + MZ

Since M is midpoint, WM = MZ, so WZ = 2 * WM

In triangle WXM, right-angled at M, with hypotenuse WX = 26, angle at X is 26°.

So, sin(angle at X) = opposite/hypotenuse = WM / WX

So sin(26°) = WM / 26

Thus WM = 26 * sin(26°)

But we need exact value? Probably not, but in the problem, likely they expect us to use the values without trig, or perhaps I missed something.

Notice that in triangle WXM, we have angles 26°, 64°, 90°, and side WX=26.

But 26 is the length, and angle is 26 degrees — coincidence?

Perhaps we can find WM using Pythagoras if we had another side, but we don't.

Another way: perhaps use the fact that in triangle WXY, we have sides and angles, but we need WZ.

Note that WZ is not directly in the triangles, but we can find coordinates or something.

Perhaps the problem expects us to realize that with x=4, and the values, but for WZ, we need to calculate.

But let's see the questions:

3. Find YZ → which is ZY = 8x+2 = 34

4. Find WX = 10x-14 = 26

5. Find YW = WY = ZY = 34 (since Y on perp bisector)

6. Find WZ

For WZ, as above, WZ = 2 * WM

In triangle WXM, WM = WX * sin(∠WXM) = 26 * sin(26°)

But sin(26°) is not nice number. Perhaps I made a mistake.

Earlier I assumed that ∠WXY = ∠ZYX implies x=4, but is that justified?

Let me go back.

We have from symmetry:

∠WXY = ∠ZXY (let's call this α)

∠WYX = ∠ZYX (call this β)

Given α = 5x+6, β = 7x-2

In triangle WXY, angles are α at X, β at Y, and γ at W, with α + β + γ = 180

Also, since the figure is symmetric, and XY is straight, the angle at W might be related.

But more importantly, consider the line WZ. At point M, the foot, we have two right triangles: WXM and WYM.

In triangle WXM, angles are: at M 90°, at X is α (since ∠WXM = ∠WXY = α), so at W is 90° - α

Similarly, in triangle WYM, angles: at M 90°, at Y is β (∠WYM = ∠WYX = β), so at W is 90° - β

Now, if X and Y are on opposite sides of WZ, then the total angle at W in triangle WXY is the sum of the two angles from the small triangles: (90° - α) + (90° - β) = 180° - α - β

But in triangle WXY, the angle at W is also 180° - α - β, from angle sum.

So it consistency checks, but doesn't give new information.

To find WZ, we need WM and MZ.

WM is in triangle WXM: WM = WX * sin(α) = (10x-14) * sin(5x+6)

Similarly, MZ = ZY * sin(β) = (8x+2) * sin(7x-2) ? No.

In triangle ZXM, since symmetric, ZM = WM, and ZX = WX, angle at X is α, so same as WXM.

But for MZ, in triangle ZXM, MZ = ZX * sin(α) = WX * sin(α) = same as WM.

So WZ = 2 * WM = 2 * WX * sin(α) = 2 * (10x-14) * sin(5x+6)

But this is messy, and we have x unknown.

Unless we can find x from another condition.

What condition have we not used? We have the side lengths, but no direct relation.

Perhaps the key is that the point Y is also on the perpendicular bisector, but we already used that for WY = ZY.

Another idea: perhaps the distance from W to X and W to Y are related, but not necessarily.

Let's look at the angles again. In the diagram, perhaps the angles at X and Y are supplementary or something, but unlikely.

Perhaps for the perpendicular bisector, the angles are such that the triangles are similar, but let's try to set the expressions.

Notice that in triangle WXY, by law of sines:

WX / sin(β) = WY / sin(α) = XY / sin(γ)

But WX = 10x-14, WY = 8x+2, α = 5x+6, β = 7x-2

So:

(10x-14) / sin(7x-2) = (8x+2) / sin(5x+6)

This is complicated, but if we assume that 5x+6 = 7x-2, then x=4, and sin(26) = sin(26), and 10*4-14=26, 8*4+2=34, so 26 / sin(26) = 34 / sin(26)? 26 = 34? No, not equal.

26 / sin(26) vs 34 / sin(26) , not equal, so law of sines not satisfied unless the sines are proportional, but here the angles are the same, so sin same, but sides different, so ratio not equal, contradiction.

Oh! So my assumption that α = β is wrong because it leads to inconsistency in law of sines.

With x=4, α=26°, β=26°, WX=26, WY=34, then in triangle WXY, side opposite to angle at Y (which is β=26°) is WX=26, side opposite to angle at X (α=26°) is WY=34.

But if angles at X and Y are both 26°, then sides opposite should be equal, but 26 ≠ 34, contradiction.

So x=4 is incorrect.

I made a mistake.

So cannot set α = β.

Back to square one.

From law of sines in triangle WXY:

WX / sin(∠WYX) = WY / sin(∠WXY)

That is:

WX / sin(β) = WY / sin(α)

Because side opposite to angle at Y is WX, side opposite to angle at X is WY.

Standard law of sines: side opposite angle.

In triangle WXY:

- Side opposite to angle at W is XY
- Side opposite to angle at X is WY
- Side opposite to angle at Y is WX

Yes.

So:

WY / sin(α) = WX / sin(β) = XY / sin(γ)

So specifically:

WY / sin(α) = WX / sin(β)

So (8x+2) / sin(5x+6) = (10x-14) / sin(7x-2)

This is an equation in x, but hard to solve analytically.

Perhaps in the context, the angles are such that 5x+6 and 7x-2 are complementary or something.

Another idea: perhaps because XY is perpendicular to WZ, and M is on it, then in the right triangles, the angles add up.

Let's consider the angle at W.

From earlier, in triangle WXM, angle at W is 90° - α

In triangle WYM, angle at W is 90° - β

If X and Y are on opposite sides of WZ, then the total angle at W for triangle WXY is (90° - α) + (90° - β) = 180° - α - β

But in triangle WXY, the angle at W is also 180° - α - β, so consistent.

To have a specific value, perhaps we can use the fact that the line XY is straight, so the angles around point M or something.

Perhaps the product or other.

Let's try to set the expression for the sides.

Recall that WX = ZX, and WY = ZY, and also, the distance between X and Y can be expressed, but complicated.

Another thought: perhaps the point M is between X and Y on the line XY, and we can find XM and YM.

In triangle WXM, XM = WX * cos(α) = (10x-14) * cos(5x+6)

In triangle WYM, YM = WY * cos(β) = (8x+2) * cos(7x-2)

Then, if X and Y are on opposite sides of M, then XY = XM + YM

If on the same side, |XM - YM|, but likely opposite sides.

So XY = XM + YM = (10x-14) cos(5x+6) + (8x+2) cos(7x-2)

But also, in triangle WXY, by law of cosines or something, but still messy.

Perhaps for the sake of this problem, they intend for us to use the property that since XY is perpendicular bisector, then for point X, WX = ZX, but we don't have ZX, and for Y, WY = ZY, which we have, but no direct link.

Let's look at the given: in the diagram, there is also the angle at X and Y, and perhaps they are related to the right angle.

Let's calculate the difference.

Suppose that the angle between WX and the perpendicular is α, and between WY and the perpendicular is β, then the angle between WX and WY is α + β if on opposite sides.

In triangle WXY, angle at W is α + β, and angles at X and Y are α and β respectively? No.

From earlier, in triangle WXY, angle at X is α, at Y is β, at W is 180 - α - β.

But from the right triangles, if X and Y are on opposite sides, angle at W is (90 - α) + (90 - β) = 180 - α - β, same thing.

So no new info.

Perhaps the key is that the line XY is the same, so the direction is the same, but for the angles to be consistent, perhaps when we consider the slope or something.

Let's try to set the expression for the side WZ.

WZ = 2 * WM

WM = WX * sin(α) = (10x-14) * sin(5x+6)

Also, from triangle WYM, WM = WY * sin(β) = (8x+2) * sin(7x-2) ? No.

In triangle WYM, angle at Y is β, so sin(β) = opposite/hypotenuse = WM / WY

So WM = WY * sin(β) = (8x+2) * sin(7x-2)

But also from triangle WXM, WM = WX * sin(α) = (10x-14) * sin(5x+6)

So we have:

(10x-14) * sin(5x+6) = (8x+2) * sin(7x-2)

This must hold.

So (10x-14) sin(5x+6) = (8x+2) sin(7x-2)

This is still hard, but perhaps for integer x, we can try values.

Let me try x=2:

Left: (20-14) sin(10+6) = 6 * sin(16°) ≈ 6*0.2756 = 1.6536

Right: (16+2) sin(14-2) = 18 * sin(12°) ≈ 18*0.2079 = 3.7422 not equal

x=3:

Left: (30-14) sin(15+6) = 16 * sin(21°) ≈ 16*0.3584 = 5.7344

Right: (24+2) sin(21-2) = 26 * sin(19°) ≈ 26*0.3256 = 8.4656 not equal

x=4:

Left: (40-14) sin(20+6) = 26 * sin(26°) ≈ 26*0.4384 = 11.3984

Right: (32+2) sin(28-2) = 34 * sin(26°) ≈ 34*0.4384 = 14.9056 not equal

x=1:

Left: (10-14) sin(5+6) = (-4) * sin(11°) ≈ -4*0.1908 = -0.7632

Right: (8+2) sin(7-2) = 10 * sin(5°) ≈ 10*0.0872 = 0.872 not equal, and negative doesn't make sense for length.

x=5:

Left: (50-14) sin(25+6) = 36 * sin(31°) ≈ 36*0.5150 = 18.54

Right: (40+2) sin(35-2) = 42 * sin(33°) ≈ 42*0.5446 = 22.8732 not equal

x=6:

Left: (60-14) sin(30+6) = 46 * sin(36°) ≈ 46*0.5878 = 27.0388

Right: (48+2) sin(42-2) = 50 * sin(40°) ≈ 50*0.6428 = 32.14 not equal

x=0:

Left: (0-14) sin(0+6) = -14 * sin(6°) ≈ -14*0.1045 = -1.463

Right: (0+2) sin(0-2) = 2 * sin(-2°) = 2*(-0.0349) = -0.0698 not equal

Perhaps x=3.5

But this is not working.

Another idea: perhaps the angles 5x+6 and 7x-2 are such that their sum is 90 degrees or something.

Set 5x+6 + 7x-2 = 90

12x +4 = 90

12x = 86

x = 86/12 = 43/6 ≈ 7.1667

Then α = 5*(43/6) +6 = 215/6 + 36/6 = 251/6 ≈ 41.833°

β = 7*(43/6) -2 = 301/6 - 12/6 = 289/6 ≈ 48.166°

Sum 90°, good.

Then WX = 10*(43/6) -14 = 430/6 - 84/6 = 346/6 = 173/3 ≈ 57.666

WY = 8*(43/6) +2 = 344/6 + 12/6 = 356/6 = 178/3 ≈ 59.333

Then WM = WX * sin(α) = (173/3) * sin(41.833°)

But also WM = WY * sin(β) = (178/3) * sin(48.166°)

Calculate sin(41.833°) ≈ sin(41.833) = ? approximately sin(42°) = 0.6694, sin(41.833) ≈ 0.667

sin(48.166°) ≈ sin(48.2) ≈ 0.745

Then left: (173/3)*0.667 ≈ 57.666*0.667 ≈ 38.46

Right: (178/3)*0.745 ≈ 59.333*0.745 ≈ 44.2 not equal.

Not good.

Set 5x+6 = 90 - (7x-2) or something.

Assume that in the right triangle, the angles are related.

Perhaps for point X, the angle α = 5x+6, and for the perpendicular, but let's think differently.

Let's look back at the diagram description. In the user's message, for problems 3-6, it says "Use the diagram below. XY is the perpendicular bisector of WZ. Find the indicated measure."

And then lists:

3. Find YZ

4. Find WX

5. Find YW

6. Find WZ

And in the diagram, likely YZ is the same as ZY = 8x+2, WX = 10x-14, YW = WY = ZY = 8x+2 (since Y on perp bisector), and WZ is to be found.

But to find WZ, we need x.

Perhaps from the angles, since XY is perpendicular to WZ, then the angle between WX and XY is α, and between WY and XY is β, and since XY is straight, the angle between WX and WY is |α - β| or α + β, but in the triangle, it's 180 - α - β, as before.

Perhaps the critical insight is that the line XY is the perpendicular bisector, so for any point on it, but specifically, the distance from W to X and from Z to X are equal, which we have, but also, the vector or something.

Another idea: perhaps the product WX * WY or something.

Let's try to set the expression for the area or other.

Perhaps in the diagram, the angles at X and Y are acute, and we can use the fact that the sum of angles in the quadrilateral or something.

Let's consider triangle WXY and triangle ZXY.

Since WX = ZX, WY = ZY, and XY common, so triangles WXY and ZXY are congruent by SSS.

Therefore, corresponding angles are equal.

So, angle at X in triangle WXY equals angle at X in triangle ZXY, which is ∠ZXY = ∠WXY = α

Similarly, angle at Y in triangle WXY equals angle at Y in triangle ZXY, which is ∠ZYX = ∠WYX = β

But in the diagram, we have ∠WXY = 5x+6, and ∠ZYX = 7x-2, and since ∠ZYX = β, and ∠WYX = β, so it's consistent, but doesn't give new equation.

For the side WZ, since M is midpoint, and in triangle WXM, etc.

Perhaps we can use the law of cosines in triangle WXY for side XY, but still.

Let's calculate the length of XY from both triangles.

From triangle WXM: XM = WX * cos(α) = (10x-14) cos(5x+6)

From triangle WYM: YM = WY * cos(β) = (8x+2) cos(7x-2)

Then if X and Y are on opposite sides of M, XY = XM + YM

If on the same side, |XM - YM|

In triangle WXY, by law of cosines, XY^2 = WX^2 + WY^2 - 2*WX*WY*cos(angle at W)

Angle at W is 180 - α - β, so cos(180 - α - β) = - cos(α + β)

So XY^2 = (10x-14)^2 + (8x+2)^2 - 2*(10x-14)*(8x+2)*(- cos(α+β)) = (10x-14)^2 + (8x+2)^2 + 2*(10x-14)*(8x+2)*cos(α+β)

On the other hand, if XY = XM + YM = (10x-14) cos(α) + (8x+2) cos(β) , assuming opposite sides.

So set equal:

[(10x-14) cos(α) + (8x+2) cos(β)]^2 = (10x-14)^2 + (8x+2)^2 + 2*(10x-14)*(8x+2)*cos(α+β)

This is very messy.

Perhaps for this problem, they intend for us to use the property that the angles are equal because of the way it's drawn, or perhaps there's a typo.

Another thought: in the diagram, perhaps ∠WXY and ∠ZYX are vertical angles or something, but unlikely.

Let's look at the values given in the diagram: in the user's initial description, for the last part, it has "5x+6" at X, "7x-2" at Y, "10x-14" for WX, "8x+2" for ZY.

Perhaps "ZY" is not WY, but in the diagram, ZY is from Z to Y, and since Y is on the perpendicular bisector, ZY = WY, so it is.

Perhaps for problem 3, YZ is the same as ZY, so 8x+2, but we need x.

Let's try to set the two expressions for WM equal.

From earlier:

WM = WX * sin(α) = (10x-14) * sin(5x+6)

WM = WY * sin(β) = (8x+2) * sin(7x-2) because in triangle WYM, sin(β) = WM / WY, so WM = WY * sin(β)

Yes.

So (10x-14) sin(5x+6) = (8x+2) sin(7x-2)

Let me denote f(x) = (10x-14) sin(5x+6) - (8x+2) sin(7x-2) = 0

Try x=3:

left: (30-14) sin(15+6) = 16 * sin(21°) ≈ 16*0.3584 = 5.7344

right: (24+2) sin(21-2) = 26 * sin(19°) ≈ 26*0.3256 = 8.4656

f(3) = 5.7344 - 8.4656 = -2.7312

x=4: 26* sin(26°) ≈ 26*0.4384 = 11.3984

34* sin(26°) = 34*0.4384 = 14.9056

f(4) = 11.3984 - 14.9056 = -3.5072

x=2: 6* sin(16°) ≈ 6*0.2756 = 1.6536

18* sin(12°) ≈ 18*0.2079 = 3.7422

f(2) = 1.6536 - 3.7422 = -2.0886

x=1: (-4)* sin(11°) ≈ -4*0.1908 = -0.7632

10* sin(5°) ≈ 10*0.0872 = 0.872

f(1) = -0.7632 - 0.872 = -1.6352

All negative, and at x=0: (-14)* sin(6°) ≈ -14*0.1045 = -1.463

2* sin(-2°) = 2*(-0.0349) = -0.0698

f(0) = -1.463 - (-0.0698) = -1.3932? No:

f(x) = left - right = [ (10x-14) sin(5x+6) ] - [ (8x+2) sin(7x-2) ]

At x=0: left = (-14) * sin(6°) ≈ -14*0.1045 = -1.463

right = (2) * sin(-2°) = 2* (-0.0349) = -0.0698

so f(0) = -1.463 - (-0.0698) = -1.3932

At x=5: left = (50-14) sin(25+6) = 36 * sin(31°) ≈ 36*0.5150 = 18.54

right = (40+2) sin(35-2) = 42 * sin(33°) ≈ 42*0.5446 = 22.8732

f(5) = 18.54 - 22.8732 = -4.3332

Always negative, and becoming more negative, but at x=6: left = 46* sin(36°) ≈ 46*0.5878 = 27.0388

right = 50* sin(40°) ≈ 50*0.6428 = 32.14

f(6) = 27.0388 - 32.14 = -5.1012

Perhaps for larger x, but sin is bounded.

Try x=3.2

α = 5*3.2+6 = 16+6=22°
β = 7*3.2-2 = 22.4-2=20.4°
WX = 10*3.2-14 = 32-14=18
WY = 8*3.2+2 = 25.6+2=27.6
left = 18 * sin(22°) ≈ 18*0.3746 = 6.7428
right = 27.6 * sin(20.4°) ≈ 27.6*0.3486 = 9.62136
f=6.7428 - 9.62136 = -2.87856

Still negative.

Perhaps I have the formula wrong.

In triangle WYM, angle at Y is β, so sin(β) = opposite/hypotenuse = WM / WY, so WM = WY * sin(β) , yes.

In triangle WXM, sin(α) = WM / WX, so WM = WX * sin(α) , yes.

So should be equal.

Unless the angle is not acute, but in diagram likely acute.

Perhaps for point Y, the angle β is not the angle in the right triangle.

Let's double-check the diagram interpretation.

In the user's message, for the last diagram, it has "5x+6" at vertex X, "7x-2" at vertex Y, "10x-14" on side WX, "8x+2" on side ZY.

And XY is perpendicular bisector of WZ.

So, at vertex X, the angle between WX and XY is 5x+6.

At vertex Y, the angle between ZY and YX is 7x-2.

Since XY is the same line, and WZ is perpendicular to it at M.

From W, the line to X makes angle α with XY, so in the right triangle WXM, angle at X is α, so yes.

From W, the line to Y makes angle with XY. But the angle given is at Y between ZY and YX.

ZY is from Z to Y, and since Z is the reflection of W, and Y on the perpendicular bisector, then the line ZY is the reflection of WY, so the angle that ZY makes with YX should be the same as the angle that WY makes with YX, because of reflection.

So ∠ between ZY and YX = ∠ between WY and YX = β

So in triangle WYM, angle at Y is β, so sin(β) = WM / WY, so WM = WY * sin(β)

Same as before.

Perhaps the issue is that for triangle WYM, if Y is on the other side, the angle might be measured differently, but usually it's the acute angle.

Perhaps in the diagram, the angle at Y is for the triangle, but let's assume that the expression is correct, and try to solve the equation.

Set (10x-14) sin(5x+6) = (8x+2) sin(7x-2)

Let me write it as:

\frac{10x-14}{8x+2} = \frac{ sin(7x-2)}{ sin(5x+6) }

Left side: (10x-14)/(8x+2) = 2(5x-7)/2(4x+1) = (5x-7)/(4x+1)

So (5x-7)/(4x+1) = sin(7x-2) / sin(5x+6)

Now, perhaps for x=2, left = (10-7)/(8+1) = 3/9 = 1/3 ≈ 0.333

right = sin(12°)/sin(16°) ≈ 0.2079/0.2756 ≈ 0.754 not equal

x=3: left = (15-7)/(12+1) = 8/13 ≈ 0.6154

right = sin(19°)/sin(21°) ≈ 0.3256/0.3584 ≈ 0.908 not equal

x=4: left = (20-7)/(16+1) = 13/17 ≈ 0.7647

right = sin(26°)/sin(26°) = 1 not equal

x=5: left = (25-7)/(20+1) = 18/21 = 6/7 ≈ 0.8571

right = sin(33°)/sin(31°) ≈ 0.5446/0.5150 ≈ 1.057 not equal

x=1: left = (5-7)/(4+1) = (-2)/5 = -0.4

right = sin(5°)/sin(11°) ≈ 0.0872/0.1908 ≈ 0.457 not equal

x=0: left = (0-7)/(0+1) = -7

right = sin(-2°)/sin(6°) ≈ (-0.0349)/0.1045 ≈ -0.334 not equal

x=6: left = (30-7)/(24+1) = 23/25 = 0.92

right = sin(40°)/sin(36°) ≈ 0.6428/0.5878 ≈ 1.093 not equal

x=7: left = (35-7)/(28+1) = 28/29 ≈ 0.9655

right = sin(47°)/sin(41°) ≈ 0.7317/0.6561 ≈ 1.115 not equal

x=8: left = (40-7)/(32+1) = 33/33 = 1

right = sin(54°)/sin(46°) ≈ 0.8090/0.7193 ≈ 1.125 not equal

At x=8, left=1, right>1.

At x=4, left=13/17≈0.764, right=1

So perhaps between x=4 and x=8, but at x=4, right=1, left<1, at x=8, left=1, right>1, so maybe at some point equal.

Try x=5: left=18/21≈0.857, right= sin(33)/sin(31)≈0.5446/0.5150≈1.057 >0.857

x=6: left=23/25=0.92, right= sin(40)/sin(36)≈0.6428/0.5878≈1.093 >0.92

x=7: left=28/29≈0.9655, right= sin(47)/sin(41)≈0.7317/0.6561≈1.115 >0.9655

x=8: left=1, right= sin(54)/sin(46)≈0.8090/0.7193≈1.125 >1

Always right > left for x>4, and for x<4, at x=3, left=8/13≈0.615, right= sin(19)/sin(21)≈0.3256/0.3584≈0.908 >0.615

At x=2, left=3/9=0.333, right= sin(12)/sin(16)≈0.2079/0.2756≈0.754 >0.333

At x=1, left= -2/5= -0.4, right= sin(5)/sin(11)≈0.0872/0.1908≈0.457 > -0.4

So always right > left for x>0, and at x=0, left= -7, right≈ -0.334, so left < right.

Perhaps for x<0, but lengths may be negative.

Try x= -1:

left = ( -10-14) sin(-5+6) = (-24) * sin(1°) ≈ -24*0.0175 = -0.42

right = ( -8+2) sin(-7-2) = (-6) * sin(-9°) = (-6)*(-0.1564) = 0.9384

f= -0.42 - 0.9384 = -1.3584

Not zero.

Perhaps I have a fundamental mistake.

Let's read the problem again.

In the user's message: "Use the diagram below. XY is the perpendicular bisector of WZ. Find the indicated measure."

Then for 3. Find YZ

4. Find WX

5. Find YW

6. Find WZ

And in the diagram, likely YZ is the length from Y to Z, which is given as 8x+2, but perhaps it's not, or perhaps for YZ, it's the same.

Perhaps "ZY" in the diagram is not the side, but in the text, it's "8x+2" for ZY, which is the same as YZ.

Another idea: perhaps for point Y, the angle 7x-2 is not ∠ZYX, but something else, but in the description, it's "7x-2" at Y, and "ZY" is labeled, so likely.

Perhaps the perpendicular bisector means that XY is perpendicular to WZ and bisects it, so M is midpoint, and XY ⊥ WZ, but X and Y are points on the line, so the line is XY.

Perhaps in the diagram, X and Y are the same point, but unlikely.

Let's look for online or standard problems, but since I can't, perhaps for this context, they intend for us to set the angles equal or use the side lengths directly.

Perhaps from the perpendicular bisector, the distance from W to X equals distance from Z to X, which is given, but also, the angle at X for the two triangles are equal, which we have.

Let's try to use the fact that in triangle WXY and triangle ZXY, since congruent, then angle at W equals angle at Z, etc.

But for WZ, perhaps we can find it as 2 * WM, and WM can be found from the area or something.

Perhaps the product.

Let's calculate the length using coordinates.

Place point M at origin (0,0), WZ on x-axis, so W at (-p, 0), Z at (p, 0), so WZ = 2p.

XY is y-axis, since perpendicular at M.

So X is at (0, a), Y is at (0, b), for some a,b.

Then WX = distance from W(-p,0) to X(0,a) = sqrt(p^2 + a^2) = 10x-14

Similarly, ZY = distance from Z(p,0) to Y(0,b) = sqrt(p^2 + b^2) = 8x+2

Also, angle at X: angle between WX and XY.

XY is the y-axis, from X(0,a) to Y(0,b), so direction vector (0, b-a)

Vector from X to W: (-p, -a) (since W(-p,0), X(0,a), so vector X to W: (-p -0, 0-a) = (-p, -a)

Vector from X to Y: (0-0, b-a) = (0, b-a)

So angle at X between vectors XW and XY.

Vectors from X: to W: (-p, -a), to Y: (0, b-a)

So cos of angle = dot product / (magnitudes)

Dot product = (-p)*(0) + (-a)*(b-a) = -a(b-a)

Magnitude of XW = sqrt(p^2 + a^2) = WX = 10x-14

Magnitude of XY = |b-a|

So cos(α) = [ -a(b-a) ] / [ (10x-14) * |b-a| ]

This is messy with absolute value.

The angle α = 5x+6 is the angle between the lines, so perhaps the acute angle.

In the right triangle, from X to M to W, with M(0,0), X(0,a), W(-p,0), so vector XM = (0,-a) if a>0, but let's assume a>0, b<0 or something.

Assume that X is above, Y below, so a>0, b<0.

Then from X(0,a), to W(-p,0), vector (-p, -a)

To Y(0,b), vector (0, b-a) = (0, negative since b<0, a>0, so b-a<0)

So the angle at X between points W, X, Y.

The vector XW = (-p, -a), vector XY = (0, b-a)

Dot product = (-p)(0) + (-a)(b-a) = -a(b-a)

Since b<0, a>0, b-a<0, so -a(b-a) = -a * negative = positive, since a>0.

Magnitude XW = sqrt(p^2 + a^2)

Magnitude XY = |b-a| = a-b since a>0>b, a-b>0

So cos(α) = [ -a(b-a) ] / [ sqrt(p^2 + a^2) * (a-b) ] = [ -a(b-a) ] / [ WX * (a-b) ]

But b-a = -(a-b), so -a * [-(a-b)] / [ WX * (a-b) ] = [ a (a-b) ] / [ WX * (a-b) ] = a / WX

So cos(α) = a / WX

Similarly, sin(α) = p / WX, because in the right triangle, opposite to α is WM = p, hypotenuse WX, so sin(α) = p / WX

Yes! So in general, for point X, sin(α) = WM / WX = p / WX

Similarly, for point Y, in triangle WYM, but for Y, the angle