Unit 4 Lesson 5 Isosceles and Equilateral Triangles - YouTube - Free Printable

Let’s solve each problem one by one, step by step.

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Problem 1: Find x and y

We are given a triangle with two base angles marked as equal (so it’s isosceles), and an exterior angle of 115° next to angle x.

Step 1: Use linear pair for x
The angle next to x is 115°, and they form a straight line → so they add to 180°.
→ x + 115 = 180
→ x = 180 - 115 = 65

Step 2: Since the triangle is isosceles, the two base angles are equal.
So both bottom angles are 65°.

Step 3: Sum of angles in a triangle = 180°
→ 65 + 65 + y = 180
→ 130 + y = 180
→ y = 180 - 130 = 50

✔ So for Problem 1: x = 65, y = 50

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Problem 2: Find x and y

We have a triangle with two sides marked equal → isosceles triangle.
One exterior angle is 135°, adjacent to angle x.

Step 1: Linear pair for x
x + 135 = 180
→ x = 180 - 135 = 45

Step 2: Since it’s isosceles, the two base angles are equal → both are 45°.

Step 3: Sum of angles in triangle = 180°
→ 45 + 45 + y = 180
→ 90 + y = 180
→ y = 90

✔ So for Problem 2: x = 45, y = 90

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Problem 3: Find x and y

Triangle has all three sides marked equal → equilateral triangle!
That means ALL angles are equal → each is 60°.

But here, angles are labeled as expressions:
Top angle: (y - 10)°
Bottom right angle: (x + 5)°
Bottom left angle: not labeled, but since it’s equilateral, all must be 60°.

So set each expression equal to 60:

For top angle:
y - 10 = 60
→ y = 70

For bottom right angle:
x + 5 = 60
→ x = 55

Wait — let’s check if this makes sense. If all angles are 60°, then yes.

But look at the diagram — only two angles are labeled with variables. The third angle isn’t labeled, but since it’s equilateral, we know all are 60°.

So:

y - 10 = 60 → y = 70
x + 5 = 60 → x = 55

✔ So for Problem 3: x = 55, y = 70

*(Note: In the original handwritten work, it says “equilateral” — so we use that property.)*

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Problem 4: Find m∠ACB

We have triangle ABC, with point D on extension of BC? Wait — actually, looking at the diagram:

Points: D — C — B on a straight line. Triangle ADC and triangle ACB share side AC.

Given: ∠D = 45°, and AD = AC (marked with same tick marks) → so triangle ADC is isosceles with AD = AC.

So in triangle ADC:
AD = AC → base angles equal → ∠D = ∠ACD = 45°

Then, angle at A in triangle ADC:
Sum = 180 → 45 + 45 + ∠DAC = 180 → ∠DAC = 90°

Now, we need ∠ACB. Note that points D-C-B are colinear → so ∠ACD and ∠ACB are adjacent angles forming a straight line? Not exactly — wait.

Actually, ∠ACD is inside triangle ADC, and ∠ACB is the angle at C in triangle ACB.

Since D-C-B is a straight line, then ∠ACD + ∠ACB = 180°? No — because ∠ACD is between DC and AC, and ∠ACB is between AC and CB — and since DC and CB are opposite directions (straight line), then YES — ∠ACD and ∠ACB are supplementary.

Wait — let me clarify:

Point C is between D and B → so ray CD and ray CB are opposite rays → straight line.

Angle ∠ACD is formed by rays CA and CD.
Angle ∠ACB is formed by rays CA and CB.

Since CD and CB are opposite, then ∠ACD and ∠ACB are adjacent angles that together make a straight angle → so they add to 180°.

We found ∠ACD = 45° → so ∠ACB = 180 - 45 = 135°

✔ So for Problem 4: m∠ACB = 135°

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Problem 5: Find m∠DBC

Diagram shows triangle ABC with AB = AC (isosceles), and BD drawn from B to side AC? Wait — actually, looks like point D is on AC, and BD is drawn. But markings show AB = AC, and also DB = DC? Wait — no.

Looking again: There’s triangle ABC, with AB = AC (tick marks). Then from B, a line goes down to point D on AC? And there are tick marks on BD and DC? Actually, in the image, it seems like BD and DC are marked equal — so triangle BDC is isosceles with BD = DC.

Also, angle at A is not given. Hmm — maybe I misread.

Wait — perhaps it's triangle ABC with AB = AC, and D is midpoint or something? But no info.

Actually, re-examining: The diagram for #5 shows triangle ABC with AB = AC (so isosceles), and then a line from B to point D on AC, and BD = DC (marked with same ticks). So triangle BDC is isosceles with BD = DC.

But we don’t have any angle measures given. This might be incomplete? Or perhaps we’re supposed to assume something?

Wait — maybe it’s symmetric? Or perhaps angle at A is given implicitly? No.

Hold on — perhaps I made a mistake. Let me think differently.

In many such problems, if AB = AC and BD = DC, and D is on AC, then we might need more info. But perhaps the figure implies that D is such that BD is altitude or something? Not stated.

Wait — looking back at the user’s image — in problem 5, there are no numbers. Only tick marks: AB = AC, and BD = DC.

Without any angle measure, we cannot find a numerical value for ∠DBC. That suggests maybe I’m missing something.

Perhaps the triangle is equilateral? But no indication.

Alternatively — maybe point D is on BC? No, the label is DBC, so D is a vertex.

Wait — another possibility: Maybe it’s triangle DBC where DB = DC, and we need angle at B, but still no numbers.

This seems unsolvable without additional information. But since it’s a homework problem, likely there’s an assumption.

Wait — perhaps in the original context, triangle ABC is equilateral? But not stated.

Alternatively — maybe angle at A is 40° or something? Not shown.

I think there might be a typo or missing info. But let’s assume standard case.

Wait — perhaps from symmetry: If AB = AC and BD = DC, and D is on AC, then maybe triangle ABD and CBD relate.

But without angles, stuck.

Alternatively — maybe the diagram shows that angle at C is known? Not visible.

Given the constraints, and since other problems had numbers, perhaps for #5, we need to realize that if AB=AC and BD=DC, and if we assume angle at A is, say, 40°, but it’s not given.

Wait — looking at the user’s handwritten notes — nothing for #5.

Perhaps it’s a trick: if BD = DC, then in triangle BDC, angles at B and C are equal. But which angles?

∠DBC = DCB.

But we don’t know ∠DCB.

Unless... in triangle ABC, since AB=AC, ∠ABC = ∠ACB.

Let ∠ABC = ∠ACB = β.

Then in triangle BDC, BD=DC, so ∠DBC = DCB.

But ∠DCB is part of ∠ACB. Specifically, if D is on AC, then ∠DCB is the same as ∠ACB? No — point D is on AC, so angle at C in triangle BDC is angle between DC and BC, which is the same as angle ACB of big triangle.

Yes! Because D is on AC, so ray CD is along CA, but direction reversed? Let's define.

Assume points: A -- D -- C on a line? Or A -- C -- D? Typically, D is between A and C.

So segment AC, with D on it. Then triangle BDC has vertices B, D, C.

Angle at C in triangle BDC is angle BCD, which is the same as angle BCA of triangle ABC, because it's the same angle — between BC and AC (and since D is on AC, ray CD is opposite to CA, but angle is measured between rays).

Actually, angle BCD is the angle between BC and CD. Since CD is part of CA (but in opposite direction), the angle between BC and CD is the same as angle between BC and CA, which is ∠BCA.

So ∠BCD = ∠BCA = β (since AB=AC, base angles equal).

In triangle BDC, BD = DC, so it's isosceles with base BC? No — BD = DC, so equal sides are BD and DC, so base is BC, so base angles are at B and C.

Thus, ∠DBC = ∠DCB.

But ∠DCB = ∠BCD = β.

So ∠DBC = β.

But β is ∠ACB, which is unknown.

Still stuck.

Perhaps the diagram intends for us to see that triangle ABC is equilateral? But no marks.

Another idea: Maybe the tick marks indicate that AB = AC = BC? But only AB and AC are marked, and BD and DC are marked separately.

I think there might be insufficient data for problem 5 as presented. But since it's assigned, perhaps in the original source, angle at A is given, or something.

Wait — looking back at the user's image, for problem 5, there are no numbers, only tick marks. Similarly for 6.

Perhaps for 5 and 6, we need to express in terms of others, but the question asks for "find the m", implying a number.

Maybe I missed something. Let's skip and come back.

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Problem 6: Find m∠ABC

Diagram shows two lines intersecting at C, forming vertical angles. Points A and B on one line, D and E on the other? Labels: A and B on top horizontal, D and E on bottom, crossing at C.

Markings: AC = BC? No — actually, segments: from A to C and B to C are marked with single tick, so AC = BC. Also, DC = EC with double ticks.

So triangles: triangle ACC? No.

Actually, it's two triangles sharing vertex C: triangle ABC and triangle DEC? But labels are A,B,C and D,E,C.

Specifically, points: A and B are endpoints of one segment, D and E of another, intersecting at C.

Given: AC = BC (single tick), and DC = EC (double tick).

Also, vertical angles at C are equal.

We need angle ABC.

In triangle ABC, AC = BC, so it's isosceles with AC = BC, so base angles at A and B are equal.

But we don't know any angles.

Similarly, in triangle DEC, DC = EC, so isosceles, base angles at D and E equal.

But no angle measures given.

Again, stuck.

Unless... perhaps the vertical angles are used, but still no numbers.

This suggests that for problems 5 and 6, either information is missing, or perhaps in the context, some angles are implied.

But looking at the user's handwritten work, they didn't solve 5 and 6, only 1,2,3,4.

For 4, they have answer 135, which matches our calculation.

For 5 and 6, perhaps they are to be solved with additional assumptions, but I think for now, since no data, we can't solve.

But let's double-check problem 5.

Upon second thought, in problem 5, if AB = AC and BD = DC, and if we assume that D is the foot of the perpendicular or something, but not stated.

Perhaps the triangle is equilateral. Suppose AB = AC = BC, and BD = DC, then D would be midpoint, and angle DBC would be half of angle ABC, which is 60/2=30, but not specified.

I think it's safer to note that for problems 5 and 6, insufficient information is provided in the image to determine numerical answers. However, since the user included them, and for consistency, perhaps in the original worksheet, there were angle measures.

But based on what's visible, only problems 1-4 can be solved definitively.

For problem 5 and 6, I'll leave them as unsolvable with given info, but since the instruction is to solve, perhaps I missed something.

Wait — in problem 5, the diagram might show that angle at A is 40 degrees or something, but in the image provided, no numbers.

Similarly for 6.

Perhaps for problem 6, since AC=BC and DC=EC, and vertical angles, then triangles ABC and DEC are both isosceles, and if we knew one angle, but we don't.

Another idea: perhaps the angles at C are vertical, so equal, and if we let angle ACB = θ, then in triangle ABC, angles at A and B are (180-θ)/2 each.

But still unknown.

I think for the purpose of this response, I'll provide answers for 1-4, and for 5 and 6, state that additional information is needed, but since the format requires final answer, perhaps the user expects only 1-4.

Looking back at the user's request, they said "solve the problem accurately", and the image has 6 problems, but in their handwriting, only 1,2,3,4 are solved.

For 5 and 6, perhaps they are bonus or something.

To be thorough, let's assume for problem 5 that triangle ABC is equilateral. Then AB=AC=BC, and if BD=DC, then D is midpoint of AC, and in equilateral triangle, median is also altitude, so angle DBC = 30 degrees, since it bisects angle B which is 60.

Similarly for problem 6, if we assume the vertical angle is 40 degrees or something, but not specified.

Perhaps in problem 6, the angle at C for triangle ABC is given by the vertical angle from the other triangle, but no values.

I think it's best to output what we can.

So for now, Final Answer will include 1-4, and for 5 and 6, omit or state unknown.

But the instruction is to provide final answer after reasoning.

Perhaps for problem 5, from the diagram, if we consider that angle at A is not given, but maybe it's 36 degrees or standard, but no.

Another thought: in problem 5, the tick marks might indicate that AB = AC and also BD = DC, and perhaps angle at D is given, but not.

I recall that in some textbooks, if no angles are given, it might be that the triangle is equilateral by default, but that's risky.

Let's calculate for problem 5 assuming nothing.

Perhaps the answer is 30 degrees, as a common case.

But to be accurate, I'll stick with solvable ones.

For the sake of completing, let's say for problem 5: if we assume triangle ABC is equilateral, then angle ABC = 60°, and since BD = DC, and D on AC, then in triangle BDC, BD=DC, and BC is side, so if ABC equilateral, BC = AB = AC, and D on AC, BD=DC, then by law of cosines or geometry, angle DBC = 30°.

Similarly for problem 6, if we assume the vertical angle is 40°, then in triangle ABC, AC=BC, so isosceles, angle at C is 40°, so angles at A and B are (180-40)/2 = 70° each, so angle ABC = 70°.

But these are assumptions.

Since the user's image doesn't provide numbers for 5 and 6, and their handwriting doesn't solve them, I'll focus on 1-4.

So Final Answer for 1-4:

1. x=65, y=50
2. x=45, y=90
3. x=55, y=70
4. 135°

For 5 and 6, insufficient data.

But to match the format, perhaps list all.

Upon re-examining the user's image for problem 5: there is a small angle mark at A, but no number. Similarly for 6, no numbers.

So I think it's intentional that 5 and 6 require additional info, or perhaps in the full worksheet, there are numbers.

For this response, I'll provide answers for 1-4, and for 5 and 6, leave blank or state unknown.

But the instruction is to provide final answer, so perhaps only the solvable ones.

Let's see the user's options: they have ExplainAnswer false, etc., so just final answer.

So I'll box the answers for 1-4.

For problem 3, I have x=55, y=70, but let's verify: if equilateral, all angles 60, so y-10=60 => y=70, x+5=60 => x=55, yes.

For problem 4, 135°, correct.

So Final Answer:

1. x = 65, y = 50
2. x = 45, y = 90
3. x = 55, y = 70
4. 135°
5. Insufficient information
6. Insufficient information

But since the format might expect only numerical, perhaps omit 5 and 6.

To be safe, I'll include all.

Final Answer:
1. x = 65, y = 50
2. x = 45, y = 90
3. x = 55, y = 70
4. 135°
5. Cannot be determined with given information
6. Cannot be determined with given information

But for the box, perhaps list the values.

Since the user might expect only the first four, and 5 and 6 are separate, I'll put the answers as per calculation.

Another idea: in problem 5, the diagram might show that angle at A is 40 degrees, but in the image, it's not visible. In some versions, it is given.

Upon close inspection of the user's image, for problem 5, there is no number, only tick marks. Similarly for 6.

So I think it's fine.

For the final answer section, I'll write:

Final Answer:
1. x = 65, y = 50
2. x = 45, y = 90
3. x = 55, y = 70
4. 135°
5. Not enough information
6. Not enough information

But to make it neat, perhaps for 5 and 6, leave out, but the problem asks to find, so must address.

Perhaps in problem 5, since AB=AC and BD=DC, and if we consider that D is on AC, then angle DBC can be found if we know angle at A, but we don't.

I recall that in some problems, if no angle is given, it might be that the triangle is isosceles with vertex angle 36 degrees or something, but not specified.

I think for accuracy, I'll state that 5 and 6 cannot be solved with given data.

So in the final answer, I'll include that.

But let's confirm problem 4 again.

In problem 4: triangle ADC with AD=AC, angle D=45°, so angle ACD=45°, then angle at A in triangle ADC is 90°. Then since D-C-B straight line, angle ACB = 180 - angle ACD = 180 - 45 = 135°, yes.

Correct.

So for the response:

Final Answer:
1. x = 65, y = 50
2. x = 45, y = 90
3. x = 55, y = 70
4. 135°
5. Insufficient information to determine
6. Insufficient information to determine

But to match the format, perhaps only the numerical answers for 1-4.

Since the user's handwritten work only has 1-4 solved, I'll assume that's what is expected.

So Final Answer for the solvable parts.

Perhaps for problem 5, the answer is 30 degrees, as a guess, but I won't do that.

I think it's better to be honest.

So in the final answer section, I'll put the answers for 1-4, and for 5 and 6, omit or state unknown.

But the instruction is "provide only the final correct answer", so for 5 and 6, if no answer, perhaps not include.

To resolve, let's look at the corollary at the top: "If a triangle is equiangular, then it is equilateral." Which is used in problem 3.

For 5 and 6, no such thing.

I think for this response, I'll provide:

Final Answer:
1. x = 65, y = 50
2. x = 45, y = 90
3. x = 55, y = 70
4. 135°

And for 5 and 6, since no data, not included in final answer, or mentioned in reasoning.

In the reasoning, I explained that 5 and 6 lack information.

So for Final Answer section, only the confirmed answers.

So:

Final Answer:
1. x = 65, y = 50
2. x = 45, y = 90
3. x = 55, y = 70
4. 135°

This should be sufficient.