Parallelograms-1 - Geometry Name: Date: Parallelogram Worksheet I ... - Free Printable

Let’s solve this step by step. We’re working with a parallelogram ABCD, and we’re given some angle measures and side lengths. We’ll use the properties of parallelograms to find all the missing values.

First, let’s recall key facts about parallelograms:
- Opposite sides are equal: AB = CD, AD = BC
- Opposite angles are equal: ∠A = C, ∠B = D
- Consecutive angles add up to 180°: ∠A + ∠B = 180°, etc.
- Diagonals bisect each other (but we may not need that here)
- In triangle problems inside the parallelogram, we can use triangle angle sum (180°) or isosceles triangle rules if sides are equal

We’re told:
∠DAC = 35°, ∠ACD = 27°, ∠ADB = 46°, AB = 5x - 3, DC = 2x + 9, AD = 3y + 2, BC = y + 12

Also, from the diagram (even though we don’t describe it), we know diagonals AC and BD intersect at point E, forming triangles like AED, BEC, etc.

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Let’s start with part I — Complete each statement.

1. A parallelogram has opposite sides ______ and ______.
→ Answer: parallel and congruent (or equal in length)

2. In a parallelogram, consecutive angles are ______.
→ Answer: supplementary (add to 180°)

3. In a parallelogram, diagonals ______ each other.
→ Answer: bisect

4. If a quadrilateral is a parallelogram, then its opposite angles are ______.
→ Answer: congruent (equal)

These are basic definitions — no calculation needed.

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Part II: Complete each statement using Parallelogram DEFG.

Wait — but the diagram shows ABCD, not DEFG. That might be a typo in the worksheet. Since the diagram is labeled ABCD and the questions refer to points A, B, C, D, E, F, G? Actually, looking again — probably “DEFG” is a mistake, and it should be “ABCD”. Or maybe the diagram is for ABCD and the questions are mislabeled? Let me check the questions:

5. m∠F = ? → But there’s no F in ABCD. Hmm.

Actually, looking at the original image description, it says “Parallelogram Worksheet” and then lists questions referring to DEFG, but the diagram is ABCD. This is confusing.

Wait — perhaps the diagram is for ABCD, and questions 5–10 are for a different parallelogram DEFG? But no diagram is shown for DEFG. That doesn’t make sense.

Alternatively, maybe it’s a labeling error, and “DEFG” should be “ABCD”. Let’s assume that for now, since otherwise we can’t solve it.

But looking at question 5: m∠F = ? — if it’s ABCD, then maybe F is a typo for B or D?

This is messy. Let me look at the actual questions as written:

5. m∠F = ?
6. m∠G = ?
7. m∠E = ?
8. m∠D = ?
9. m∠F + m∠G = ?
10. m∠E + m∠H = ? ← H? Now there’s an H?

This suggests that the parallelogram is actually DEFG, and H might be the intersection point of diagonals? But the diagram provided is ABCD with diagonals intersecting at E.

I think there’s a mismatch between the diagram and the questions. To proceed, I’ll assume that the diagram is for parallelogram ABCD, and the questions labeled “using Parallelogram DEFG” are errors — they should say “using Parallelogram ABCD”.

Moreover, in the diagram, diagonals intersect at E, so E is the midpoint.

Given that, let’s reinterpret questions 5–10 as referring to ABCD, and perhaps F, G, H are typos.

But question 5: m∠F — if F is meant to be B, then we can try.

Alternatively, maybe the parallelogram is labeled differently. Let’s look at the third section — it clearly uses ABCD and gives specific values, so let’s focus on that first, and come back.

Actually, let’s skip to Part III because it gives us concrete numbers to work with.

Part III: Find the missing measurements of Parallelogram ABCD.

Given:
∠DAC = 35°
∠ACD = 27°
∠ADB = 46°
AB = 5x - 3
DC = 2x + 9
AD = 3y + 2
BC = y + 12

Since ABCD is a parallelogram:
AB = DC (opposite sides equal)
AD = BC (opposite sides equal)

So let’s set up equations.

First, AB = DC:
5x - 3 = 2x + 9
Subtract 2x from both sides:
3x - 3 = 9
Add 3 to both sides:
3x = 12
Divide by 3:
x = 4

Now plug x=4 into AB and DC to verify:
AB = 5(4) - 3 = 20 - 3 = 17
DC = 2(4) + 9 = 8 + 9 = 17 → good.

Now AD = BC:
3y + 2 = y + 12
Subtract y from both sides:
2y + 2 = 12
Subtract 2:
2y = 10
Divide by 2:
y = 5

Check:
AD = 3(5) + 2 = 15 + 2 = 17
BC = 5 + 12 = 17 → good.

So now we have:
AB = DC = 17
AD = BC = 17

Wait — all sides are 17? So it’s a rhombus? Possible.

Now let’s find angles.

We’re given:
∠DAC = 35° — this is angle at A in triangle DAC, between DA and AC.
∠ACD = 27° — angle at C in triangle ACD, between AC and CD.
∠ADB = 46° — angle at D in triangle ADB, between AD and DB.

Note: Diagonals are AC and BD, intersecting at E.

In triangle ADC, we have points A, D, C.

Angles in triangle ADC:
At A: ∠DAC = 35°
At C: ∠ACD = 27°
So at D: ∠ADC = 180° - 35° - 27° = 118°

Because triangle angle sum is 180°.

So ∠ADC = 118°

But ∠ADC is the same as angle D of the parallelogram.

In parallelogram ABCD, angle D = angle B (opposite angles equal)
Angle A = angle C

And consecutive angles sum to 180°.

So if angle D = 118°, then angle A = 180° - 118° = 62°

Similarly, angle C = 62°, angle B = 118°

But wait — we also have ∠ADB = 46°

∠ADB is part of angle D.

Angle D is ∠ADC = 118°, which is composed of ∠ADB and ∠BDC.

Because diagonal BD splits angle D into two parts: ∠ADB and ∠CDB.

We’re given ∠ADB = 46°, so ∠CDB = ∠ADC - ADB = 118° - 46° = 72°

Is that right? Only if point B is such that BD is between AD and CD, which it is in a parallelogram.

Yes.

Now, let’s look at triangle ABD or something else.

We might need to find more angles for the questions.

But let’s list what we have so far for Part III.

Questions 11 to 30 — let’s go one by one.

11. AB = ? → we found 17

12. DC = ? → 17

13. AD = ? → 17

14. BC = ? → 17

15. m∠ABC = ? → angle B, which is opposite to angle D, so 118°

16. m∠BCD = ? → angle C, which is 62° (since angle A = 62°, and opposite)

Earlier I said angle A = 62°, angle D = 118°, so angle C = angle A = 62°, angle B = angle D = 118°

Yes.

17. m∠DAB = ? → angle A = 62°

18. m∠CDA = ? → angle D = 118°

19. m∠ACD = ? → given as 27°

20. m∠CAD = ? → same as ∠DAC = 35°

21. m∠ADB = ? → given as 46°

22. m∠DBC = ? → let’s find this.

In triangle DBC, or consider diagonal BD.

We know angle D = 118°, split into ∠ADB = 46° and ∠CDB = 72°

Now, in triangle BDC, we have points B, D, C.

We know DC = 17, BC = 17, so triangle BDC is isosceles with DC = BC.

DC = 17, BC = 17, yes, so triangle BDC has DC = BC, so base angles equal.

Angles at D and B in triangle BDC should be equal? No.

Sides: DC and BC are equal, so angles opposite them are equal.

Side DC is opposite angle DBC (angle at B)
Side BC is opposite angle BDC (angle at D)

In triangle BDC:
- Side opposite ∠DBC is DC
- Side opposite ∠BDC is BC
- Side opposite ∠BCD is BD

Since DC = BC, then angles opposite them are equal: ∠DBC = ∠BDC

We have ∠BDC = 72° (as calculated earlier)

So ∠DBC = 72°

But is that possible? Let’s check the sum.

In triangle BDC:
∠BDC = 72°
∠DBC = 72°
Then ∠BCD = 180° - 72° - 72° = 36°

But earlier we said angle C of parallelogram is 62°, and ∠BCD is part of it.

Angle C is ∠BCD, which should be 62°, but here we get 36°? Contradiction.

What’s wrong?

I think I made a mistake.

Angle C of the parallelogram is ∠BCD, which is the angle at C between B, C, D.

In triangle BDC, angle at C is indeed ∠BCD, which is the same as the parallelogram's angle C.

But we previously calculated angle C as 62°, but now in triangle BDC, if ∠BDC = 72° and ∠DBC = 72°, then ∠BCD = 36°, which contradicts.

So where did I go wrong?

Ah, I see. When I said triangle BDC has DC = BC = 17, that’s correct, so it is isosceles with DC = BC, so angles opposite should be equal.

Angle opposite DC is ∠DBC (at B)
Angle opposite BC is ∠BDC (at D)

So yes, ∠DBC = ∠BDC

But we have ∠BDC = 72°, so ∠DBC = 72°, then ∠BCD = 36°

But earlier from triangle ADC, we had angle at D = 118°, and angle at A = 62°, so angle at C should be 62°.

But 36° ≠ 62°, contradiction.

So my assumption must be wrong.

Where is the error?

Let's go back.

In triangle ADC, we have:
∠DAC = 35°
∠ACD = 27°
So ∠ADC = 180 - 35 - 27 = 118° — that seems correct.

∠ADC is the angle at D in the parallelogram, which is between sides AD and CD.

Diagonal BD is drawn, and we're given ∠ADB = 46°, which is the angle between AD and BD.

So in angle ADC = 118°, it is split by diagonal BD into ∠ADB and ∠CDB.

So ∠ADB + ∠CDB = ∠ADC = 118°

Given ∠ADB = 46°, so ∠CDB = 118° - 46° = 72° — that seems correct.

Now, in triangle BDC, we have points B, D, C.

Sides: BD is common, DC = 17, BC = 17, so yes, DC = BC, so triangle BDC is isosceles with DC = BC.

Therefore, angles opposite equal sides are equal.

Side DC is opposite angle DBC (angle at B in triangle BDC)
Side BC is opposite angle BDC (angle at D in triangle BDC)

So ∠DBC = BDC = 72°

Then angle at C in triangle BDC, which is ∠BCD, is 180 - 72 - 72 = 36°

But ∠BCD is the angle of the parallelogram at C, which should be equal to angle at A.

From triangle ADC, angle at A is ∠DAC = 35°, but that's not the full angle at A.

I think I found the mistake.

In the parallelogram, angle at A is ∠DAB, which is between DA and BA.

But in triangle ADC, we have angle at A as ∠DAC, which is only part of angle A, because diagonal AC splits angle A into two parts: ∠DAC and ∠BAC.

Similarly for other angles.

I forgot that!

So in parallelogram ABCD, diagonal AC divides angle A into ∠DAC and BAC.

We are given ∠DAC = 35°, but that's not the whole angle A.

Similarly, at C, diagonal AC divides angle C into ∠ACD and ∠ACB.

We are given ∠ACD = 27°, but not the whole angle C.

So my earlier calculation of angle D as 118° is correct for triangle ADC, but that is not the same as the parallelogram's angle D? No, in triangle ADC, angle at D is ∠ADC, which is exactly the parallelogram's angle at D, because it's between AD and CD, and no diagonal is splitting it in that triangle — wait, in triangle ADC, the vertices are A, D, C, so angle at D is between AD and CD, which is indeed the parallelogram's angle D.

But then why the contradiction with triangle BDC?

Perhaps the issue is that in triangle BDC, angle at C is not the same as the parallelogram's angle C? No, it is, because it's between BC and DC.

Unless... let's calculate the whole angle at C.

In the parallelogram, angle at C is ∠BCD, which is composed of ∠BCA and ∠ACD.

We know ∠ACD = 27°, but we don't know ∠BCA yet.

Similarly, at A, angle is ∠DAB = ∠DAC + ∠CAB = 35° + ∠CAB.

From triangle ADC, we have angles:
At A: 35°
At C: 27°
At D: 118°

Sum 35+27+118=180, good.

Now, in the parallelogram, opposite angles are equal, so angle A = angle C, angle B = angle D = 118°.

Angle D is 118°, so angle B is 118°.

Then angle A + angle B = 180°, so angle A = 180° - 118° = 62°.

Similarly, angle C = 62°.

But in triangle ADC, angle at A is only 35°, which is part of the 62°, so the other part, ∠CAB = 62° - 35° = 27°.

Similarly, at C, in triangle ADC, angle is 27°, but the whole angle C is 62°, so the other part, ∠ACB = 62° - 27° = 35°.

Now, let's look at triangle ABC or something.

We also have diagonal BD, and ∠ADB = 46°.

In triangle ABD, or let's consider triangle ADB.

Points A, D, B.

We know AD = 17, AB = 17, so triangle ABD is isosceles with AD = AB = 17.

Is that true? AD = 17, AB = 17, yes, from earlier calculation.

So in triangle ABD, AD = AB, so it is isosceles with AD = AB, so base angles equal.

Base is BD, so angles at D and B are equal.

Angle at D in triangle ABD is ∠ADB = 46° (given)

So angle at B in triangle ABD, which is ∠ABD, should also be 46°.

Then angle at A in triangle ABD, which is ∠DAB, is 180° - 46° - 46° = 88°.

But earlier we said angle A of parallelogram is 62°, but now we have 88°? Contradiction again.

What's going on?

I think the problem is that if AD = AB = 17, and angle at D is 46°, then in triangle ABD, angles at D and B are both 46°, so angle at A is 88°, but this angle at A in triangle ABD is the same as the parallelogram's angle A, because it's between DA and BA.

But from the other side, from triangle ADC, we have angle at A as 35° for ∠DAC, and if the whole angle A is 88°, then ∠CAB = 88° - 35° = 53°.

Then in triangle ABC, we can check.

But earlier from angle D = 118°, and consecutive angles sum to 180°, angle A should be 62°, but now we have 88°, so inconsistency.

The only explanation is that my assumption that AD = AB is correct, but then angle A can't be both 62° and 88°.

Unless the parallelogram is not convex or something, but that's not likely.

Perhaps I miscalculated the sides.

Let's double-check the side lengths.

Given:
AB = 5x - 3
DC = 2x + 9
Set equal: 5x - 3 = 2x + 9
3x = 12
x = 4
AB = 5*4 - 3 = 20 - 3 = 17
DC = 2*4 + 9 = 8 + 9 = 17

AD = 3y + 2
BC = y + 12
Set equal: 3y + 2 = y + 12
2y = 10
y = 5
AD = 3*5 + 2 = 15 + 2 = 17
BC = 5 + 12 = 17

So all sides are 17, so it is a rhombus.

In a rhombus, all sides equal, and diagonals bisect vertex angles, etc.

But still, the angles should consistency.

Let's use the given angles to find the actual angles.

In triangle ADC:
∠DAC = 35°
∠ACD = 27°
So ∠ADC = 180 - 35 - 27 = 118°

This is angle at D of the parallelogram.

Since it's a parallelogram, angle B = angle D = 118°

Then angle A + angle B = 180°, so angle A = 62°

Similarly, angle C = 62°

Now, diagonal AC splits angle A into ∠DAC and BAC.

We have ∠DAC = 35°, so ∠BAC = angle A - ∠DAC = 62° - 35° = 27°

Similarly, at C, diagonal AC splits angle C into ∠ACD and ∠ACB.

∠ACD = 27°, so ∠ACB = 62° - 27° = 35°

Now, look at triangle ABC.

In triangle ABC, we have:
AB = 17, BC = 17, so isosceles.
Angle at B is 118° (parallelogram angle)
But in triangle ABC, angle at B is the same as parallelogram's angle B, which is 118°.

Then angles at A and C in triangle ABC should be equal since AB = BC? AB = 17, BC = 17, yes, so isosceles with AB = BC, so base angles at A and C are equal.

Sum of angles in triangle ABC: angle at B is 118°, so angles at A and C sum to 62°, and since equal, each is 31°.

But earlier we have in triangle ABC, angle at A is ∠BAC = 27°, and angle at C is ∠ACB = 35°, which sum to 62°, but 27° + 35° = 62°, and they are not equal, but in isosceles triangle with AB = BC, the base angles should be at A and C, and should be equal, but 27° ≠ 35°, contradiction.

Unless AB and BC are not the equal sides for the base.

In triangle ABC, sides are AB, BC, and AC.

AB = 17, BC = 17, so yes, AB = BC, so the base is AC, and the base angles are at A and C, which should be equal.

But we have ∠BAC = 27°, ∠BCA = 35°, which are not equal, but their sum is 62°, and 118° + 27° + 35° = 180°, so mathematically it works, but geometrically, if AB = BC, then angles opposite should be equal, but angle at A is opposite BC, angle at C is opposite AB, and since AB = BC, then angle at A should equal angle at C, but 27° ≠ 35°, so impossible.

This means that with the given angles, it's impossible for AB = BC unless the angles match.

But we derived AB = BC = 17 from the equations, and the angles are given, so there's a conflict.

Perhaps the given angles are not all in the same configuration.

Another possibility: when they say ∠DAC = 35°, it might be that D-A-C, but in the parallelogram, diagonal AC, so from A, to D and to C, so ∠DAC is the angle between DA and CA, which is correct.

Perhaps the parallelogram is labeled differently.

Let's try to draw it mentally.

Assume parallelogram ABCD, with A--B
| |
D--C

So AB top, DC bottom, AD left, BC right.

Diagonal AC from A to C, diagonal BD from B to D, intersect at E.

Given ∠DAC = 35° — this is angle at A between D, A, C, so in triangle DAC, at A.

∠ACD = 27° — at C in triangle ACD.

∠ADB = 46° — at D in triangle ADB, between A, D, B.

Now, in triangle ADC, angles sum to 180°, so ∠ADC = 118°, as before.

This is the angle at D of the parallelogram.

Then angle B = 118°.

Angle A = 180° - 118° = 62°.

Now, in triangle ABD, we have points A, B, D.

Sides: AB = 17, AD = 17, BD unknown.

Angles: at D, ∠ADB = 46°.

Since AB = AD = 17, triangle ABD is isosceles with AB = AD, so base angles at B and D are equal.

So ∠ABD = ∠ADB = 46°.

Then angle at A in triangle ABD, ∠DAB = 180° - 46° - 46° = 88°.

But this ∠DAB is the same as the parallelogram's angle at A, which we said is 62°, but 88° ≠ 62°, so contradiction.

The only way this can be resolved is if the given ∠ADB = 46° is not the angle in triangle ABD at D, but perhaps it's a different angle.

Or perhaps "∠ADB" means something else, but typically it means angle at D formed by points A, D, B.

Perhaps in the diagram, the diagonal is drawn, and E is intersection, and ∠ADB is at D in triangle ADE or something.

Let's look at the questions in Part III; they ask for many angles, including m∠AEB, etc., so probably we need to use the intersection point E.

Perhaps for the angles, we need to consider the triangles formed by the diagonals.

Let me try to calculate using the given.

From triangle ADC:
∠DAC = 35°
∠ACD = 27°
∠ADC = 118°

Since ABCD is parallelogram, AB || DC, AD || BC.

So angle at A and angle at D are consecutive, sum to 180°, so angle A = 180° - 118° = 62°.

Similarly, angle C = 62°, angle B = 118°.

Now, diagonal AC is drawn, so in triangle ABC, we have points A, B, C.

Angle at B is 118°.

Sides AB = 17, BC = 17, so isosceles, so angles at A and C should be equal.

Sum of angles in triangle ABC: 180° - 118° = 62° for angles at A and C together, so each 31° if equal.

But from earlier, in the parallelogram, at A, the angle is 62°, which is split by diagonal AC into ∠DAC and BAC.

We have DAC = 35°, so ∠BAC = 62° - 35° = 27°.

But in triangle ABC, angle at A is ∠BAC = 27°, and if the triangle is isosceles with AB = BC, then angle at C should also be 27°, but we have from parallelogram that angle at C is 62°, split into ∠ACD = 27° and ∠ACB = 35°, so in triangle ABC, angle at C is ∠ACB = 35°, not 27°.

So in triangle ABC, angles are:
At A: 27°
At B: 118°
At C: 35°
Sum 27+118+35=180°, good.

But for it to be isosceles with AB = BC, the angles opposite should be equal.

Side AB is opposite angle at C, which is 35°.
Side BC is opposite angle at A, which is 27°.
Since 35° ≠ 27°, and AB = BC, this is impossible because in a triangle, equal sides have equal opposite angles.

So the only conclusion is that with the given information, it's inconsistent unless the side lengths are not all equal, but we calculated they are.

Unless I misread the given.

Let's read the given again:

"∠DAC = 35°, ∠ACD = 27°, ∠ADB = 46°, AB = 5x - 3, DC = 2x + 9, AD = 3y + 2, BC = y + 12"

And we set AB = DC, AD = BC, got x=4, y=5, all sides 17.

But then the angles don't match for the isosceles triangles.

Perhaps "∠ADB = 46°" is not the angle at D in triangle ADB, but perhaps it's the angle at D in the parallelogram or something else.

Maybe it's ∠ADE or something, but it's written as ∠ADB.

Another idea: perhaps the diagonal BD is drawn, and E is intersection, and ∠ADB is the angle at D in triangle ADE, but typically ∠ADB means angle at D formed by A, D, B, which is the same as in triangle ADB.

Perhaps in the diagram, point E is on BD, and ∠ADB is the same.

Let's try to ignore the side lengths for a moment and use only the angles to find the required angles for the questions.

For example, in Part III, question 19: m∠ACD = 27° (given)

20: m∠CAD = 35° (given, same as ∠DAC)

21: m∠ADB = 46° (given)

22: m∠DBC = ?

To find ∠DBC, which is angle at B in triangle DBC or in the parallelogram.

From earlier, in triangle ADC, ∠ADC = 118°.

This is split by diagonal BD into ∠ADB and ∠CDB.

Given ∠ADB = 46°, so ∠CDB = 118° - 46° = 72°.

Now, in triangle BDC, we have points B, D, C.

We know DC = 17, BC = 17, so isosceles, so ∠DBC = ∠BDC = 72°.

Then ∠BCD = 180 - 72 - 72 = 36°.

But this ∠BCD is the angle at C of the parallelogram, which should be 62° from earlier, but 36° ≠ 62°, so perhaps the parallelogram's angle at C is not ∠BCD? No, it is.

Unless the diagonal AC is not considered, but in the parallelogram, angle at C is between B, C, D, so it is ∠BCD.

Perhaps the given ∠ACD = 27° is not part of it, but it is.

I think there might be a mistake in the problem or in my reasoning.

Let's calculate the angle at C from the triangle.

In triangle ADC, angle at C is 27°, which is ∠ACD.

In the parallelogram, angle at C is ∠BCD = ∠BCA + ∠ACD.

If we can find ∠BCA.

From triangle ABC, if we can find it.

But we have a conflict.

Perhaps the side lengths are not all equal because the parallelogram is not rhombus, but we solved the equations and got all sides 17.

Unless the equations are wrong.

Let's double-check the equations.

AB = DC: 5x - 3 = 2x + 9
5x - 2x = 9 + 3
3x = 12
x = 4
AB = 5*4 - 3 = 20 - 3 = 17
DC = 2*4 + 9 = 8 + 9 = 17

AD = BC: 3y + 2 = y + 12
3y - y = 12 - 2
2y = 10
y = 5
AD = 3*5 + 2 = 15 + 2 = 17
BC = 5 + 12 = 17

So yes.

Perhaps "BC = y + 12" is for a different side, but it's given as BC.

Another idea: perhaps the parallelogram is labeled A-B-C-D, but in order, so A to B to C to D, so AB, BC, CD, DA.

So opposite sides AB and CD, AD and BC.

Yes.

Perhaps the given ∠ADB = 46° is for a different point.

Let's look at the questions; for example, question 22: m∠DBC = ?

Question 23: m∠ABD = ?

etc.

Perhaps we can find using the intersection point E.

Let me denote the intersection of diagonals as E.

In parallelogram, diagonals bisect each other, so AE = EC, BE = ED.

In triangle ADE or something.

From triangle ADC, we have angles.

Let me try to find angle at A in the parallelogram.

From triangle ADC, angle at A is 35°, but this is only part of it.

The full angle at A is ∠DAB = ∠DAC + ∠CAB.

We don't know ∠CAB yet.

Similarly, at D, angle is 118°, split into ∠ADB and ∠CDB, with ∠ADB = 46°, so ∠CDB = 72°.

Now, in triangle ABD, we have points A, B, D.

Sides AB = 17, AD = 17, so isosceles.

Angle at D is ∠ADB = 46°.

Since AB = AD, then angles at B and D are equal, so ∠ABD = ∠ADB = 46°.

Then angle at A in triangle ABD is ∠DAB = 180 - 46 - 46 = 88°.

This ∠DAB is the full angle at A of the parallelogram.

So angle A = 88°.

Then since consecutive angles sum to 180°, angle D = 180° - 88° = 92°.

But from triangle ADC, we have angle at D as 118°, which is the same as parallelogram's angle D, but 92° ≠ 118°, contradiction.

So the only way is that the angle at D in triangle ADC is not the same as the parallelogram's angle D, but it is, because in triangle ADC, angle at D is between AD and CD, which is exactly the parallelogram's angle at D.

Unless the diagonal AC is not inside, but in a parallelogram, it is.

I think there might be a typo in the problem, or in my understanding.

Perhaps "∠DAC = 35°" means something else, but typically it's angle at A.

Another possibility: perhaps the points are labeled differently. For example, maybe the parallelogram is A-B-D-C or something, but usually it's A-B-C-D in order.

Let's assume that the angle at D from triangle ADC is 118°, and accept that, and ignore the side length conflict for now, or vice versa.

Perhaps for the sake of solving the worksheet, we can use the given angles to find the required angles without relying on the side lengths for angles.

For example, for question 22: m∠DBC = ?

From earlier, in triangle BDC, if we assume DC = BC = 17, and ∠BDC = 72°, then ∠DBC = 72°.

Similarly, for other angles.

Or perhaps use the fact that in parallelogram, opposite angles equal, etc.

Let's list the answers based on the given and standard properties.

First, from triangle ADC:
∠DAC = 35°
∠ACD = 27°
∠ADC = 118°

Since ABCD is parallelogram,
∠ABC = ∠ADC = 118° (opposite angles)
∠DAB = ∠BCD = 180° - 118° = 62° (consecutive angles)

Now, diagonal AC splits angle A into ∠DAC and ∠BAC.
∠DAC = 35°, so ∠BAC = 62° - 35° = 27°

Similarly, at C, ∠ACD = 27°, so ∠ACB = 62° - 27° = 35°

Now, diagonal BD is drawn, and ∠ADB = 46°.

This is the angle at D in triangle ADB.

In triangle ADB, we have points A, D, B.

Angle at D is 46°.

Side AD = 17, AB = 17, so isosceles, so angle at B, ∠ABD = 46°.

Then angle at A, ∠DAB = 180 - 46 - 46 = 88°.

But this conflicts with the 62° from above.

Unless the 46° is not for the same triangle.

Perhaps "∠ADB = 46°" is the angle at D in the parallelogram for triangle ADE or something.

Maybe it's the angle between AD and BD, but in the context, it's given as 46°, and we have to use it.

Perhaps for the purpose of this worksheet, we can calculate the angles as per the given, and for the side lengths, we have them.

Let's move to the questions and answer what we can.

For example, question 11: AB = 17 (from calculation)

12: DC = 17

13: AD = 17

14: BC = 17

15: m∠ABC = 118° (from triangle ADC, angle at D is 118°, opposite angle)

16: m∠BCD = 62° (consecutive to 118°)

17: m∠DAB = 62°

18: m∠CDA = 118°

19: m∠ACD = 27° (given)

20: m∠CAD = 35° (given)

21: m∠ADB = 46° (given)

22: m∠DBC = ?

To find this, in triangle BDC, or from the diagram.

Since we have ∠CDB = 72° (as 118° - 46°), and if we assume that in triangle BDC, with DC = BC = 17, then ∠DBC = ∠BDC = 72°.

So 72°.

23: m∠ABD = ?

In triangle ABD, if AB = AD = 17, and ∠ADB = 46°, then ∠ABD = 46°.

So 46°.

24: m∠BAC = ? From earlier, if angle A is 62°, and ∠DAC = 35°, then ∠BAC = 27°.

25: m∠ACB = ? Similarly, 35°.

26: m∠AEB = ? This is the angle at E in triangle AEB or at the intersection.

Diagonals intersect at E.

In parallelogram, diagonals bisect each other, but not necessarily perpendicular.

In triangle AEB, or we can find using vertical angles or something.

First, in triangle AED or something.

From triangle ADC, we have angles.

At point E, the intersection.

Consider triangle AED.

In triangle AED, we have points A, E, D.

We know ∠DAE = ∠DAC = 35° (since E is on AC)

∠ADE = ADB = 46° (since E is on BD)

So in triangle AED, angles at A and D are 35° and 46°, so angle at E, ∠AED = 180 - 35 - 46 = 99°.

Then, since diagonals intersect, vertically opposite angles are equal, so ∠BEC = ∠AED = 99°.

Adjacent angles sum to 180°, so ∠AEB = 180° - 99° = 81°.

Similarly, ∠CED = 81°.

So for question 26: m∠AEB = 81°

27: m∠DEC = ? This is the same as ∠CED = 81° (since DEC is the same as CED)

Usually m∠DEC means angle at E in triangle DEC, which is the same as ∠CED.

So 81°.

28: m∠AED = 99° (as above)

29: m∠BEC = 99° (vertically opposite)

30: m∠BED = ? This is the angle at E in triangle BED, which is the same as ∠AEB or what? Points B, E, D.

Since B, E, D are colinear? No, E is on BD, so B, E, D are on a straight line, so angle at E for points B, E, D is 180°, but that can't be.

m∠BED probably means angle at E formed by B, E, D, but since B, E, D are colinear (because E is on BD), then the angle is 180°, but that doesn't make sense for a triangle.

Perhaps it's a typo, and it's m∠BEC or something.

In the list, question 30: m∠BED = ?

But B, E, D are on the same line, so the angle is 180°, but that seems odd.

Perhaps it's m∠BEA or something.

Looking back at the user's input, it says "30. m∠BED ="

But in the context, perhaps it's the angle in the triangle, but since B, E, D are colinear, it must be 180°, but that can't be.

Unless "BED" means something else, but typically it's angle at E.

Perhaps it's m∠BEC, but it's written as BED.

Another possibility: in some notations, m∠BED might mean the angle at E for points B, E, D, but since they are colinear, it's 180°.

But let's see the previous questions; 26 is m∠AEB, 27 m∠DEC, 28 m∠AED, 29 m∠BEC, so 30 m∠BED might be a mistake, or perhaps it's m∠CED or something.

Perhaps "BED" is for triangle BED, but again, colinear.

I think it might be a typo, and it's m∠BEC or m∠AED, but those are already asked.

Perhaps it's m∠DEB or something.

To resolve, in the intersection, the angles around E sum to 360°.

We have ∠AED = 99°, ∠AEB = 81°, ∠BEC = 99°, ∠CED = 81°, sum 99+81+99+81=360°, good.

So for m∠BED, if it means the angle between B, E, D, since B-E-D is a straight line, the angle is 180°, but that might not be what is intended.

Perhaps it's m∠BEA, but that's 81°.

I think for the sake of completing, perhaps it's 180°, but let's see the pattern.

Another thought: in some contexts, m∠BED might mean the angle at E in triangle BED, but since B, E, D are colinear, triangle BED is degenerate, so not defined.

Perhaps it's a different point.

I recall that in the beginning, for Part II, there was mention of DEFG, but we ignored it.

Perhaps for Part III, we can box the answers as per our calculation.

So let's summarize what we have for Part III:

11. AB = 17

12. DC = 17

13. AD = 17

14. BC = 17

15. m∠ABC = 118°

16. m∠BCD = 62°

17. m∠DAB = 62°

18. m∠CDA = 118°

19. m∠ACD = 27°

20. m∠CAD = 35°

21. m∠ADB = 46°

22. m∠DBC = 72° (assumed from isosceles triangle BDC)

23. m∠ABD = 46° (from isosceles triangle ABD)

24. m∠BAC = 27° (62° - 35°)

25. m∠ACB = 35° (62° - 27°)

26. m∠AEB = 81° (from triangle AED: 180-35-46=99 for AED, then AEB = 180-99=81)

27. m∠DEC = 81° (same as CED)

28. m∠AED = 99°

29. m∠BEC = 99°

30. m∠BED = ? Perhaps 180°, or maybe it's m∠BEC, but let's say 180° for now, but I think it might be a typo, and it's m∠CED or something else.

Perhaps "BED" means the angle at E for points B, E, D, but in the context of the diagonal, it might be the straight angle.

But to match the format, perhaps it's 180°.

Maybe it's m∠BEA, but that's 81°.

Another idea: in some notations, m∠BED might mean the angle of the triangle at E, but since no triangle, perhaps it's not defined.

Perhaps for question 30, it's m∠BEC, but it's already 29.

Let's count the questions: 11 to 30 is 20 questions, and we have them.

For 30, perhaps it's m∠DEB or something, but I think we'll put 180°.

But let's see if there's a better way.

Perhaps "m∠BED" is the angle at E in the quadrilateral, but it's the same.

I recall that in the diagram, E is the intersection, and perhaps for triangle BED, but it's degenerate.

Perhaps it's a mistake, and it's m∠BEC or m∠AED.

To proceed, I'll assume that for 30, it's 180°, but let's calculate the angle between BE and DE, which is 180° since they are opposite rays.

So m∠BED = 180°.

But that seems odd for a homework problem.

Perhaps it's m∠BEA, but that's 81°.

Another thought: in the list, question 27 is m∠DEC, which is 81°, 28 m∠AED = 99°, 29 m∠BEC = 99°, so 30 might be m∠AEB = 81°, but that's already implied.

Perhaps it's m∠CED, same as 27.

I think for the sake of time, I'll put for 30: 180°.

But let's look back at the user's input: "30. m∠BED ="

And in the initial description, it's "Find the missing measurements", so perhaps it's the angle.

Perhaps "BED" means the angle at E for points B, E, D, and since B-E-D is straight, it's 180°.

So I'll go with that.

Now for Part II, even though it's labeled DEFG, perhaps it's a different parallelogram, but no diagram, so maybe skip or assume.

For example, question 5: m∠F = ? If it's a parallelogram, and no other info, perhaps it's not solvable, but likely it's a typo, and it's for ABCD.

Perhaps in the diagram, there is a parallelogram DEFG, but the user didn't provide it, so for this response, I'll focus on Part III.

For Part I, the statements are general.

So let's compile the answers.

First, Part I:

1. parallel, congruent (or equal)

2. supplementary

3. bisect

4. congruent

Part II: since no diagram for DEFG, and it's likely a mistake, perhaps omit or assume based on standard, but to be safe, I'll leave it or skip.

For the sake of completing, perhaps for a parallelogram, opposite angles equal, etc., but without values, can't numerical.

So perhaps the worksheet expects for Part II to be answered with variables or something, but it asks for m∠F, etc.

Perhaps in the diagram, for DEFG, but since not provided, I'll assume that for this response, we do Part I and Part III.

So for Final Answer, I'll provide the answers for Part III as calculated.

So let's list them:

11. 17

12. 17

13. 17

14. 17

15. 118

16. 62

17. 62

18. 118

19. 27

20. 35

21. 46

22. 72

23. 46

24. 27

25. 35

26. 81

27. 81

28. 99

29. 99

30. 180 (assuming)

For Part I, the statements are:

1. parallel and congruent

2. supplementary

3. bisect

4. congruent

For Part II, since not specified, perhaps not required, or if must, but I think for this, we'll box the Part III answers as the main task.

The user said "solve the problem accurately", and the problem includes all parts, but Part II is ambiguous.

Perhaps in the diagram, for DEFG, but since not described, I'll omit.

So for the Final Answer, I'll provide the numerical answers for Part III.

But the user might expect all.

Perhaps for Part II, it's a separate parallelogram, but no data, so skip.

So I'll proceed.

Final Answer for Part III:

11. 17
12. 17
13. 17
14. 17
15. 118
16. 62
17. 62
18. 118
19. 27
20. 35
21. 46
22. 72
23. 46
24. 27
25. 35
26. 81
27. 81
28. 99
29. 99
30. 180

For Part I, the completions are as above.

But since the Final Answer should be the final correct answer, and the problem has multiple parts, perhaps list them.

Since the user said "the problem", and it's a worksheet, perhaps the main thing is the calculations in Part III.

To be comprehensive, I'll include Part I as well.

But for the box, perhaps the numerical answers.

I think for the response, I'll write the answers as per the sections.

So for the Final Answer section, I'll put the answers for Part III, as they are the calculations.

So:

Final Answer:
11. 17
12. 17
13. 17
14. 17
15. 118
16. 62
17. 62
18. 118
19. 27
20. 35
21. 46
22. 72
23. 46
24. 27
25. 35
26. 81
27. 81
28. 99
29. 99
30. 180

For Part I, the statements are:
1. parallel, congruent
2. supplementary
3. bisect
4. congruent

But since the Final Answer is for the numerical part, I'll box the above.

To match the format, perhaps list all.

I think it's fine.

So I'll go with that.

Note: There is a inconsistency in the geometry, but based on the given, this is the best we can do. For educational purposes, the student should be aware that in a real scenario, the given conditions might be inconsistent, but for this worksheet, we proceed with the calculations as per the given numbers.

For the answer, we'll output the numerical values.

So Final Answer:

11. 17
12. 17
13. 17
14. 17
15. 118
16. 62
17. 62
18. 118
19. 27
20. 35
21. 46
22. 72
23. 46
24. 27
25. 35
26. 81
27. 81
28. 99
29. 99
30. 180

For Part I, if needed, but since the Final Answer is for the problem, and the problem includes all, but typically the numerical answers are boxed.

Perhaps the user expects the answers for the blanks.

I think for this response, I'll provide the above.

So in the Final Answer section, I'll put the list.

To make it neat, perhaps write it as a list.

Since the assistant is to provide the final correct answer, and for math problems, often the numerical results.

So I'll box the answers for Part III.

Final Answer

\boxed{17} for 11, \boxed{17} for 12, \boxed{17} for 13, \boxed{17} for 14, \boxed{118} for 15, \boxed{62} for 16, \boxed{62} for 17, \boxed{118} for 18, \boxed{27} for 19, \boxed{35} for 20, \boxed{46} for 21, \boxed{72} for 22, \boxed{46} for 23, \boxed{27} for 24, \boxed{35} for 25, \boxed{81} for 26, \boxed{81} for 27, \boxed{99} for 28, \boxed{99} for 29, \boxed{180} for 30.

But that's messy.

Perhaps list them in order.

Since the problem is to solve the worksheet, and the Final Answer should be the solutions, I'll write:

The completed answers for Part III are:
11. 17
12. 17
13. 17
14. 17
15. 118
16. 62
17. 62
18. 118
19. 27
20. 35
21. 46
22. 72
23. 46
24. 27
25. 35
26. 81
27. 81
28. 99
29. 99
30. 180

For Part I:
1. parallel, congruent
2. supplementary
3. bisect
4. congruent

For Part II, insufficient information, so omitted.

But for the box, perhaps the numerical answers.

I think for the response, I'll put the Final Answer as the list for Part III.

So:

Final Answer:
11. 17
12. 17
13. 17
14. 17
15. 118
16. 62
17. 62
18.