Quadrilateral proofs worksheet with answers: Fill out & sign ... - Free Printable

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

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Problem 1:

We are given that ABCD is a parallelogram. The student wrote a proof to show that ∠B ≅ ∠D (opposite angles are congruent).

The steps so far:

1. ABCD is a parallelogram → Given
2. BC ≅ AD and AB ≅ DC → Opposite sides of a parallelogram are congruent
3. AC ≅ CA → Reflexive Postulate (same segment)
4. △ABC ≅ △CDA → Side-Side-Side (SSS) — because all three sides match: AB≅DC, BC≅AD, AC≅CA
5. ∠B ≅ ∠D → ???

Now we need the reason for step 5.

Since we proved the two triangles are congruent (△ABC ≅ △CDA), then their corresponding parts must be congruent. That’s called CPCTC: Corresponding Parts of Congruent Triangles are Congruent.

∠B is in triangle ABC, and ∠D is in triangle CDA — and they correspond to each other in the congruent triangles.

So the correct reason is:
→ Corresponding parts of congruent triangles are congruent.

That matches option 3.

✔ Final Answer for Problem 1: 3

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Problem 2:

Given: Parallelogram ABCD with diagonal AC drawn.
Prove: △ABC ≅ △CDA

Let’s think about what we know from a parallelogram:

- Opposite sides are congruent: AB ≅ CD, BC ≅ DA
- Diagonal AC is shared → AC ≅ CA (reflexive property)

So in triangles ABC and CDA:

- AB ≅ CD (side)
- BC ≅ DA (side)
- AC ≅ CA (side)

That’s SSS (Side-Side-Side) congruence!

Alternatively, you could use SAS if you used alternate interior angles from parallel lines, but since we have all three sides matching, SSS is direct.

But wait — let’s check the diagram. In parallelogram ABCD, diagonal AC connects A to C.

Triangle ABC has sides: AB, BC, AC
Triangle CDA has sides: CD, DA, CA

Yes — AB = CD, BC = DA, AC = CA → SSS.

So the proof would be:

Statements | Reasons
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1. ABCD is a parallelogram | Given
2. AB ≅ CD, BC ≅ DA | Opposite sides of parallelogram are congruent
3. AC ≅ CA | Reflexive Property
4. △ABC ≅ △CDA | SSS Congruence

✔ So the proof is complete using SSS.

Final Answer for Problem 2:
Proof completed using SSS congruence as shown above.

*(Note: Since this is a “prove” question, there’s no multiple choice — just write the proof. But since the user didn’t ask to write it out fully, and only wants final answers where applicable, we’ll note that the triangles are congruent by SSS.)*

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Problem 3:

Given: Quadrilateral ABCD, diagonal AFEC (wait — probably typo? Should be AC?), AE ⊥ FC, BF ⊥ AC, DE ⊥ AC, ∠1 ∠2
Prove: ABCD is a parallelogram.

Wait — looking at the diagram description: Points F and E are on diagonal AC. BF and DE are perpendicular to AC. ∠1 is at B, ∠2 is at D.

Actually, re-reading: “diagonal AFEC” — that doesn’t make sense. Probably meant diagonal AC, with points F and E on it.

Assuming: Diagonal AC, with F and E on AC. BF ⊥ AC, DE ⊥ AC, and ∠1 ≅ ∠2 (where ∠1 = ∠ABF? or ∠CBF? Diagram shows ∠1 at vertex B between AB and BF; ∠2 at vertex D between CD and DE).

Also, AE ≅ FC? Wait — given says: “AE ≅ FC”, “BF ⊥ AC”, “DE ⊥ AC”, “∠1 ≅ ∠2”

Goal: Prove ABCD is a parallelogram.

Strategy: Show opposite sides are parallel or congruent, or diagonals bisect, etc.

Since BF ⊥ AC and DE ⊥ AC, then BF || DE (both perpendicular to same line).

Also, ∠1 ≅ ∠2 — these are angles at B and D.

If we can prove triangles ABF and CDE are congruent, or something similar...

Wait — perhaps better approach: Use right triangles.

Consider triangles BFA and DEC.

We know:

- BF ⊥ AC → ∠BFA = 90°
- DE ⊥ AC → ∠DEC = 90°
- ∠1 ≅ ∠2 (given)
- Also, AE ≅ FC (given)

But AE and FC are segments on AC. Let’s denote:

Let’s say points on AC: A --- F --- E --- C ? Or A --- E --- F --- C? Not clear.

Wait — given says “AE ≅ FC”. If E and F are on AC, and AE = FC, that might help.

Perhaps consider triangles ABF and CDE.

But we don’t have enough yet.

Alternative idea: Since BF and DE are both perpendicular to AC, and if we can show BF = DE, then maybe quadrilateral BEDF is a rectangle or something — not sure.

Wait — another thought: If we can show that AB || CD and AD || BC, or that opposite sides are congruent.

Use the fact that ∠1 ≅ ∠2 and the perpendiculars to create congruent triangles.

Look at triangles ABF and CDE:

- ∠AFB = ∠CED = 90° (perpendiculars)
- ∠1 ≅ ∠2 (given)
- Need one more thing — side?

Given AE ≅ FC. But AE and FC are not directly in those triangles.

Unless... suppose AC is the diagonal, and F and E are points such that AF + FE + EC = AC.

If AE ≅ FC, and assuming order A-F-E-C, then AE = AF + FE, FC = FE + EC, so AF + FE = FE + EC ⇒ AF = EC.

Oh! That’s key.

If AE ≅ FC, and points are colinear on AC in order A, F, E, C, then:

AE = AF + FE
FC = FE + EC
Set equal: AF + FE = FE + EC ⇒ AF = EC

So now we have AF ≅ EC.

Now look at triangles ABF and CDE:

- ∠AFB = ∠CED = 90°
- ∠1 ≅ ∠2 (given)
- AF ≅ EC (just deduced)

Therefore, by AAS (Angle-Angle-Side), △ABF ≅ △CDE.

Then, corresponding parts: AB ≅ CD, and BF ≅ DE.

Similarly, now look at triangles ADE and CBF? Or maybe triangles AFD and CEB?

Wait — we also have BF ≅ DE (from above), and both perpendicular to AC, so actually, since BF and DE are both perpendicular to AC and equal in length, and if we consider the distance between them...

But to prove ABCD is a parallelogram, we already have AB ≅ CD from above.

Now need to show AD ≅ BC or AB || CD.

From △ABF ≅ △CDE, we also get ∠BAF ≅ ∠DCE.

But ∠BAF and ∠DCE are alternate interior angles if we consider transversal AC cutting lines AB and CD.

If ∠BAF ≅ ∠DCE, and they are on opposite sides of transversal AC, then AB || CD.

Similarly, we can try to prove AD || BC.

Consider triangles ADE and CBF.

We have:

- DE ⊥ AC, BF ⊥ AC → ∠AED = ∠CFB = 90°
- We have AF = EC (from earlier)
- What about AE and FC? Given AE ≅ FC — yes!
Wait, AE ≅ FC is given, and we used it to get AF = EC.

Actually, AE ≅ FC is given, and we derived AF = EC.

But for triangles ADE and CBF:

- AE ≅ FC (given)
- DE ≅ BF (from previous congruence)
- ∠AED = ∠CFB = 90°

So by SAS (since right triangles, legs AE=FC, DE=BF), △ADE ≅ △CBF.

Thus, AD ≅ CB.

Now we have:

- AB ≅ CD (from first congruence)
- AD ≅ BC (from second congruence)

Therefore, opposite sides are congruent → ABCD is a parallelogram.

✔ So the proof works.

Final Answer for Problem 3:
ABCD is a parallelogram because opposite sides are congruent, proven via congruent right triangles using AAS and SAS.

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Problem 4:

Given: Quadrilateral ABCD, AD ≅ BC, ∠DAE ≅ ∠BCE. Line segments AC, DB, and FG intersect at E. Prove: △AEF ≅ △CEG

Diagram: Points F on AD, G on BC? And FG passes through E, intersection of diagonals.

Given:

- AD ≅ BC
- ∠DAE ≅ ∠BCE
- Diagonals AC and BD intersect at E
- FG is another line through E, with F on AD, G on BC? (assumed from diagram)

Need to prove △AEF ≅ △CEG

First, note that ∠DAE and ∠BCE are angles at A and C.

∠DAE is part of angle at A, between DA and AE (which is part of diagonal AC).

Similarly, ∠BCE is at C, between BC and CE (part of AC).

Since AD ≅ BC and ∠DAE ≅ ∠BCE, and if we can find another pair of angles or sides...

Notice that ∠AEF and ∠CEG are vertical angles? Because FG and AC intersect at E, so ∠AEF and ∠CEG are vertical angles → therefore congruent.

Is that true? Point E is intersection of AC, DB, and FG. So yes, lines AC and FG cross at E, so ∠AEF and ∠CEG are vertical angles → ∠AEF ≅ ∠CEG.

Now, we have:

- ∠DAE ≅ BCE (given)
- But ∠DAE is the same as ∠FAE (if F is on AD)
- Similarly, ∠BCE is same as ∠GCE (if G is on BC)

So in triangles AEF and CEG:

- ∠FAE ≅ ∠GCE (given)
- ∠AEF ≅ CEG (vertical angles)
- Now, do we have a side?

We need one side to use ASA or AAS.

What about AE and CE? Are they equal? Not necessarily — unless diagonals bisect, which we don't know.

But we have AD ≅ BC, but that's the whole sides, not parts.

Wait — perhaps use the fact that in triangles ADE and CBE?

Consider triangles ADE and CBE:

- AD ≅ BC (given)
- ∠DAE ≅ ∠BCE (given)
- ∠AED ≅ CEB? Why? Because they are vertical angles! Diagonals AC and BD intersect at E, so ∠AED and ∠CEB are vertical angles → congruent.

Therefore, by ASA: △ADE ≅ △CBE.

Then, corresponding parts: AE ≅ CE, and DE ≅ BE.

Great! So now we know AE ≅ CE.

Back to triangles AEF and CEG:

- ∠FAE ≅ ∠GCE (given)
- AE ≅ CE (just proven)
- ∠AEF ≅ ∠CEG (vertical angles)

Therefore, by ASA (Angle-Side-Angle), △AEF ≅ △CEG.

✔ Proof complete.

Final Answer for Problem 4:
△AEF ≅ △CEG by ASA congruence, using given angles, vertical angles, and AE ≅ CE from congruent triangles ADE and CBE.

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Problem 5:

Given: Parallelogram FLSH, diagonal FGAS (probably typo — should be FS? Or FA? Wait — "diagonal FGAS" — likely means diagonal FS, with points G and A on it? Diagram shows diagonal FS, with G and A on it, LG ⊥ FS, HA ⊥ FS.

Prove: △LGS ≅ △HAF

Parallelogram FLSH — so vertices F, L, S, H. Typically labeled in order, so FL || SH, LS || FH, etc.

Diagonal FS — connects F to S.

Points G and A on FS, with LG ⊥ FS, HA ⊥ FS.

So LG and HA are both perpendicular to diagonal FS.

In parallelogram FLSH:

- Opposite sides congruent: FL ≅ HS, LS ≅ FH
- Opposite angles congruent
- Diagonals bisect each other, but here we have perpendiculars from L and H to diagonal FS.

Note: Since it's a parallelogram, and we're dropping perpendiculars from L and H to diagonal FS.

Actually, in parallelogram FLSH, points L and H are opposite vertices? Let's assume standard labeling: F to L to S to H to F.

So diagonal FS connects F to S, and the other diagonal is LH.

But here, we have perpendiculars from L and H to diagonal FS.

LG ⊥ FS, HA ⊥ FS — so G and A are feet of perpendiculars from L and H to FS.

We need to prove △LGS ≅ △HAF.

Look at the triangles:

△LGS: points L, G, S
△HAF: points H, A, F

Both are right triangles, since LG ⊥ FS and HA ⊥ FS, so ∠LGS = 90°, ∠HAF = 90°.

Now, in parallelogram FLSH:

- LS ≅ FH (opposite sides)
- Also, ∠LSF ≅ ∠HFS? Not sure.

Note that since FLSH is a parallelogram, vectorially or by properties, the distance from L to FS should equal distance from H to FS? Not necessarily, unless symmetric.

But let's think differently.

Consider that in parallelogram, diagonal FS divides it into two congruent triangles: △FLS ≅ △SHF? Actually, △FLS and △HSF — but standard is △FLS ≅ △HSF by SSS or SAS.

More carefully: In parallelogram FLSH, diagonal FS creates △FLS and △HSF.

Actually, △FLS and △HSF share side FS, and FL ≅ HS, LS ≅ HF, so yes, △FLS ≅ △HSF by SSS.

Therefore, area of △FLS = area of △HSF.

But also, area of △FLS = (1/2)*FS*LG (since LG is height from L to base FS)
Area of △HSF = (1/2)*FS*HA (height from H to base FS)

Since areas equal and same base FS, then LG = HA.

So the heights are equal: LG ≅ HA.

Now, back to triangles LGS and HAF.

We have:

- LG ≅ HA (just shown)
- ∠LGS = ∠HAF = 90°
- Now, need another side or angle.

What about GS and AF? Or LS and HF?

LS and HF are opposite sides of parallelogram, so LS ≅ HF.

But in triangles LGS and HAF, sides are:

△LGS: LG, GS, LS
△HAF: HA, AF, HF

We have LG ≅ HA, LS ≅ HF, and both right-angled.

So by HL (Hypotenuse-Leg) congruence for right triangles:

- Hypotenuse LS ≅ hypotenuse HF
- Leg LG ≅ leg HA
- Right angles at G and A

Therefore, △LGS ≅ △HAF by HL.

✔ Proof complete.

Final Answer for Problem 5:
△LGS ≅ △HAF by HL congruence, since both are right triangles with congruent hypotenuses (LS ≅ HF) and congruent legs (LG ≅ HA).

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## Final Answers Summary:

Problem 1: 3
Problem 2: △ABC ≅ △CDA by SSS congruence
Problem 3: ABCD is a parallelogram (proven via congruent triangles)
Problem 4: △AEF ≅ △CEG by ASA congruence
Problem 5: △LGS ≅ △HAF by HL congruence

Since the user may expect boxed answers for multiple choice, and others as proofs, but per instruction, we provide final answer after reasoning.

For clarity, since Problem 1 is multiple choice, and others are proofs, but the format asks for final answer, I'll list them as:

Final Answer:
1. 3
2. Proven by SSS
3. Proven by showing opposite sides congruent via triangle congruence
4. Proven by ASA
5. Proven by HL

But to strictly follow "final answer" as single box, perhaps only for MCQ.

Looking back, only Problem 1 has options. Others are "prove", so no numerical answer.

But the instruction says: "provide only the final correct answer" — for proofs, it's the conclusion.

To comply, I'll write:

Final Answer:
For Problem 1: \boxed{3}
For Problem 2: \triangle ABC \cong \triangle CDA \text{ by SSS}
For Problem 3: ABCD \text{ is a parallelogram}
For Problem 4: \triangle AEF \cong \triangle CEG \text{ by ASA}
For Problem 5: \triangle LGS \cong \triangle HAF \text{ by HL}

But since the system might expect one box, and Problem 1 is the only MCQ, perhaps just box 3.

Re-reading user request: "solve the problem accurately" — and there are 5 problems.

But in the response format, it says "Final Answer: [Provide only the final correct answer]"

Perhaps for each problem, but since it's one response, I'll summarize.

Given the constraints, I'll provide the answer for Problem 1 as it's multiple choice, and for others state the result.

But to be safe, since the user might be expecting the answer to the first problem primarily (as it's listed first and has options), and others are proofs, I'll box the answer for Problem 1.

However, the instruction says "solve the problem" — plural implied.

After careful thought, I'll provide all final conclusions clearly.

Final Answer:
\boxed{3} for Problem 1.
For Problems 2-5, the proofs are complete as reasoned above, establishing the required congruences or properties.