Geometry problem from S.S.T.B. Quadrilateral Proofs worksheet, asking to prove quadrilateral ABCD is a parallelogram using given angle congruences.

Let’s solve each problem one by one, step by step. We’ll use simple logic and geometry rules you’ve learned.

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Problem 1: Complete the proof that opposite angles in a parallelogram are congruent.

We’re given: ABCD is a parallelogram.
We need to prove: ∠B ≅ ∠D (and also ∠A ≅ ∠C, but we’ll focus on ∠B and D as shown).

The table has statements and reasons — let’s fill them in logically.

Given:
- AB || CD → Statement 1
- AD || BC → Statement 2 (since it’s a parallelogram, both pairs of opposite sides are parallel)

Now, when two parallel lines are cut by a transversal, alternate interior angles are congruent.

Look at diagonal AC — it cuts across AB and CD (which are parallel). So:

→ ∠1 ≅ ∠4 (alternate interior angles) → This matches statement 3.

Similarly, diagonal AC also cuts across AD and BC (also parallel), so:

→ ∠2 ≅ ∠3 (alternate interior angles) → This should be statement 4.

Then, if ∠1 ≅ ∠4 and ∠2 ≅ ∠3, then adding them:

∠1 + ∠2 ≅ ∠3 + ∠4 → That’s statement 5.

But ∠1 + 2 = ∠BAD (angle at A), and ∠3 + ∠4 = ∠BCD (angle at C). Wait — actually, looking at the diagram labels:

In the diagram, angle B is made up of ∠1 and ∠2? Actually, no — let’s check the labeling.

Looking at the figure for Problem 1: Diagonal AC is drawn. Angles are labeled:

At vertex A: ∠1 and ∠2 (so ∠DAB = ∠1 + ∠2)
At vertex C: ∠3 and ∠4 (so ∠BCD = ∠3 + ∠4)
At vertex B: just ∠B
At vertex D: just ∠D

Wait — actually, from the proof steps, they say:

Statement 3: ∠1 ∠4 → these are alternate interior angles from AB || CD and transversal AC.

Statement 4: ??? → must be ∠2 ≅ ∠3 (from AD || BC and transversal AC)

Then Statement 5: ∠1 + 2 ≅ ∠3 + ∠4 → which means ∠DAB ≅ ∠BCD → so opposite angles at A and C are congruent.

But the question asks: “What do the reason justify that ∠B ≅ ∠D?”

Hmm — maybe there’s a mix-up. Let’s read carefully.

Actually, the proof is set up to show ∠A ≅ ∠C first (via ∠1+∠2 ≅ ∠3+∠4). But the question says: “What do the reason justify that ∠B ≅ ∠D?”

That suggests maybe the proof is incomplete or mislabeled? Or perhaps we need to do a similar thing with diagonal BD?

Wait — look again at the options for the multiple choice:

> What do the reason justify that ∠B ≅ ∠D?
> 1) Opposite angles in a parallelogram are congruent. ← this is what we’re trying to prove! Can’t use it as a reason.
> 2) Parallelograms have two congruent corresponding diagonals. ← not true generally, and not relevant here.
> 3) Corresponding parts of congruent triangles are congruent. ← possible, if we prove triangles congruent.
> 4) Alternate interior angles in congruent triangles are congruent. ← doesn’t make sense — alternate interior angles come from parallel lines, not necessarily congruent triangles.

Actually, the best answer is option 3: Corresponding parts of congruent triangles are congruent.

Why? Because if we draw diagonal AC, we can prove triangle ABC ≅ triangle CDA (by ASA or SAS), then ∠B and ∠D would be corresponding parts.

But in the current proof setup, they’re using angle addition to show ∠A ≅ ∠C. To show ∠B ≅ ∠D, we’d need to either:

- Draw diagonal BD and do similar steps, OR
- Use the fact that consecutive angles are supplementary, and since ∠A ≅ ∠C, then ∠B = 180 - ∠A, ∠D = 180 - ∠C, so ∠B ≅ ∠D.

But none of that is in the table.

Wait — perhaps the proof is meant to show ∠A ≅ ∠C, and the question is misphrased? Or maybe in the diagram, ∠B and D are being referred to via other angles?

Alternatively, maybe after proving ∠1 ≅ ∠4 and ∠2 ≅ ∠3, then since ∠ABC = ∠1 + something? No.

Let me re-express:

In parallelogram ABCD, with diagonal AC:

Triangles ABC and CDA share side AC.

AB = CD (opposite sides of parallelogram)
AD = BC (same reason)
AC = AC (common)

So by SSS, triangle ABC ≅ triangle CDA.

Therefore, ∠B ≅ ∠D (corresponding parts of congruent triangles).

Ah! So even though the table is showing angle additions for ∠A and ∠C, the actual justification for ∠B ≅ ∠D comes from triangle congruence.

So the correct answer to the multiple choice is:

3) Corresponding parts of congruent triangles are congruent.

Because once you prove the two triangles formed by a diagonal are congruent, their corresponding angles (including ∠B and ∠D) are congruent.

✔ Final Answer for Problem 1 Multiple Choice: 3

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Problem 2: Given Parallelogram ABCD with diagonal AC. Prove: ΔABC ≅ ΔCDA

This is straightforward.

In parallelogram ABCD:

- AB || CD and AD || BC (definition)
- AB ≅ CD (opposite sides of parallelogram are congruent)
- AD ≅ BC (same reason)
- AC ≅ AC (reflexive property)

So in triangles ABC and CDA:

AB ≅ CD
BC ≅ DA
AC ≅ CA

Therefore, by SSS (Side-Side-Side) congruence, ΔABC ≅ ΔCDA.

You could also use SAS:

AB ≅ CD
∠BAC ≅ ∠DCA (alternate interior angles, since AB || CD and AC is transversal)
AC ≅ CA

So SAS also works.

Either way, the triangles are congruent.

✔ Proof complete.

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Problem 3: Given Quadrilateral ABCD, diagonal BD, AB ≅ DC, AB ⊥ BC, DC ⊥ BC, ∠1 ≅ ∠2. Prove: ABCD is a parallelogram.

Given:

- AB ≅ DC
- AB ⊥ BC → so ∠ABC = 90°
- DC ⊥ BC → so ∠DCB = 90°
- ∠1 ≅ ∠2 → looking at diagram, ∠1 and 2 are probably angles at B and D with respect to diagonal BD? Wait, diagram shows diagonal BD, and ∠1 and 2 are likely ABD and ∠CDB or something.

Actually, from standard notation, if diagonal BD is drawn, and ∠1 and ∠2 are marked near B and D, likely ∠1 = ∠ABD, ∠2 = ∠CDB.

But we’re told ∠1 ≅ ∠2.

Also, AB ⊥ BC and DC ⊥ BC → so AB and DC are both perpendicular to BC → meaning AB || DC (because both perpendicular to same line).

And we’re given AB ≅ DC.

So we have one pair of opposite sides that are both parallel and congruent → therefore, ABCD is a parallelogram.

That’s a theorem: If one pair of opposite sides of a quadrilateral are both parallel and congruent, then it’s a parallelogram.

So we don’t even need ∠1 ∠2? Maybe it’s redundant, or maybe to confirm something else.

But strictly speaking, since AB || DC (both ⊥ BC) and AB ≅ DC, then ABCD is a parallelogram.

✔ Proven.

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Problem 4: In diagram below of parallelogram ABCD, BE ≅ DF and ∠BAE ≅ ∠DCF. Line segments AC, EF, and BD intersect at E. Prove: ΔAEF ≅ ΔCGE

Wait — point G? The diagram isn't fully clear, but from context, probably G is intersection point? Or typo?

Looking back: “Line segments AC, EF, and BD intersect at E.” So all three meet at E.

But then it says prove ΔAEF ≅ ΔCGE — so G must be another point. Probably G is on BD or something.

Wait — perhaps it's a typo and should be ΔAEF ≅ ΔCEF? Or maybe G is the intersection point?

Actually, rereading: “Prove: ΔAEF ≅ ΔCGE”

And given: BE ≅ DF, ∠BAE ≅ ∠DCF.

Since ABCD is a parallelogram, AB || CD, AD || BC, AB ≅ CD, AD ≅ BC.

Diagonals bisect each other, but here we have points E and F on BD? BE ≅ DF suggests E and F are on BD such that BE = DF.

Assume BD is a diagonal, E and F are points on BD with BE = DF.

Also, ∠BAE ≅ ∠DCF.

We need to prove ΔAEF ≅ ΔCGE — but who is G? Perhaps G is the intersection of AC and EF? But it says AC, EF, BD intersect at E — so E is common point.

This is confusing without the diagram.

Alternative approach: Since ABCD is parallelogram, and BE ≅ DF, and assuming E and F are on BD, then perhaps AE and CF are drawn.

Given ∠BAE ∠DCF.

Also, AB ≅ CD (parallelogram).

So in triangles ABE and CDF:

AB ≅ CD
BE ≅ DF (given)
∠BAE ∠DCF (given)

But that’s SSA, which is not a congruence criterion.

Unless the angles are included? ∠BAE is at A, between BA and AE; ∠DCF is at C, between DC and CF.

Not included between the sides we have.

Perhaps we need to consider triangles AEF and CGE.

Maybe G is point C? Typo?

Another idea: Perhaps "G" is a typo and should be "C", so prove ΔAEF ≅ ΔCEF? But that might not be true.

Or perhaps G is the midpoint or something.

Given the ambiguity, and since this is a common type, likely we can prove triangles AEB and CFD congruent first.

Wait — let’s try:

In parallelogram ABCD, AB || CD, so ∠ABE ≅ ∠CDF (alternate interior angles if we consider BD as transversal).

Is that true? AB || CD, BD transversal → yes, ∠ABD ≅ ∠CDB.

So ∠ABE ≅ ∠CDF.

Given BE ≅ DF, and AB ≅ CD.

So in triangles ABE and CDF:

AB ≅ CD
BE ≅ DF
∠ABE ≅ ∠CDF

Therefore, by SAS, ΔABE ≅ ΔCDF.

Thus, AE ≅ CF, and ∠BAE ≅ ∠DCF (already given), and ∠AEB ≅ ∠CFD.

Now, since AC and EF intersect at E, and we have AE ≅ CF, and perhaps vertical angles or something.

To prove ΔAEF ≅ ΔCGE — still stuck on G.

Perhaps G is point C, and it's ΔAEF ≅ ΔCEF? But F and E are different.

Another thought: Maybe "G" is the intersection point, but it's already E.

I think there might be a typo in the problem. Perhaps it's prove ΔAEF ≅ ΔCFG or something.

Given time, I'll assume that with the congruence of ΔABE ≅ ΔCDF, and since ABCD is parallelogram, diagonals bisect, but here E is on BD.

Perhaps after proving AE ≅ CF, and since AC is diagonal, and EF is another line, and they intersect at E, then for triangles AEF and CEF, but not sure.

Let’s skip and come back.

Actually, looking at the last part: "Prove: ΔAEF ≅ ΔCGE"

And in the diagram description, it says "line segments AC, EF, and BD intersect at E" — so E is the common intersection point.

Then G must be another point. Perhaps G is on AC or something.

Maybe G is point C, and it's a typo, should be ΔAEF ≅ ΔCEF.

But let's assume that from ΔABE ≅ ΔCDF, we have AE ≅ CF, and ∠AEB ≅ ∠CFD.

Now, ∠AEB and ∠CEF are vertical angles? If EF is a straight line, and AC crosses it at E, then ∠AEB and ∠CEF might be vertical if B-E-D and A-E-C are lines.

Assume that AC and BD intersect at some point, but the problem says they intersect at E, so E is the intersection of diagonals? In a parallelogram, diagonals bisect each other, so if E is intersection of AC and BD, then AE ≅ EC, BE ≅ ED.

But here, we have BE ≅ DF, and F is another point.

If E is intersection of diagonals, then BE = ED, so if BE ≅ DF, then DF = BE = ED, so F is beyond D or something.

This is messy.

Perhaps F is on BD extended, but let's try a different approach.

Given the complexity and potential diagram issues, and since this is for a student, I'll provide a general method.

But to move forward, let's assume that with the given, we can prove the triangles congruent using ASA or SAS.

For example, if we can show that AE ≅ CG, EF ≅ GE, and included angle, but we don't know G.

Perhaps "G" is a typo and it's "C", so prove ΔAEF ≅ ΔCEF.

Then, if E is on AC, and F is on BD, and we have AE ≅ CE (if E is midpoint), but not given.

I think there's an error in the problem statement or my understanding.

Let's look at Problem 5 instead.

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Problem 5: Given parallelogram FLSM, diagonal FGSL, LG ⊥ SH, FH ≅ FS. Prove: ΔGLS ≅ ΔHMF

First, parallelogram FLSM — so vertices F, L, S, M.

Diagonal FGSL — that doesn't make sense. Probably "diagonal FS" or "diagonal LM".

It says "diagonal FGSL" — likely a typo. Perhaps "diagonal FS" and "GL" or something.

Given: LG ⊥ SH, FH ≅ FS.

Prove: ΔGLS ≅ ΔHMF.

Points: F, L, S, M are vertices of parallelogram.

Diagonal — probably FS or LM.

LG ⊥ SH — so L to G perpendicular to S to H.

FH ≅ FS — so F to H equals F to S.

This is very unclear without diagram.

Perhaps G and H are points on the diagonals.

In many such problems, G and H are feet of perpendiculars or intersections.

Given the ambiguities in Problems 4 and 5, and since the user may have intended only the first few, I'll focus on what's clear.

But to fulfill the request, let's try to resolve Problem 4 with a reasonable assumption.

Revisiting Problem 4:

Assume that in parallelogram ABCD, diagonal BD is drawn. Points E and F are on BD such that BE ≅ DF. Also, ∠BAE ≅ ∠DCF. Diagonals AC and BD intersect at O, but the problem says AC, EF, BD intersect at E — so perhaps E is the intersection point, and F is another point on BD with BE ≅ DF.

If E is the intersection of diagonals, then in parallelogram, E is midpoint, so BE = ED.

Given BE ≅ DF, then DF = BE = ED, so F is such that D-F-B or something, with DF = BE.

Suppose B-E-D on diagonal BD, with BE = ED (since E is midpoint).

Given BE ≅ DF, so DF = BE = ED, so if D-F, then F is beyond D, with DF = ED, so E-D-F, with ED = DF.

Then F is symmetric to B with respect to D.

Then we have points A, E, F, C, G.

Prove ΔAEF ≅ ΔCGE.

Still vague.

Perhaps G is point C, and it's ΔAEF ≅ ΔCEF.

Then, with AE and CE, if E is midpoint, AE = CE.

From earlier, if we can prove ΔABE ≅ ΔCDF, then AE = CF.

Then in triangles AEF and CEF, we have AE = CF, EF common, but not sufficient.

Angles at E might be equal if vertical.

I think for the sake of time, I'll provide answers for the clear problems.

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Summary of Answers:

Problem 1 Multiple Choice:
The reason that justifies ∠B ≅ ∠D is:
3) Corresponding parts of congruent triangles are congruent.

Problem 2 Proof:
In parallelogram ABCD, AB ≅ CD, AD ≅ BC, and AC ≅ AC. By SSS, ΔABC ≅ ΔCDA.

Problem 3 Proof:
Since AB ⊥ BC and DC ⊥ BC, AB || DC. Given AB ≅ DC, so one pair of opposite sides are both parallel and congruent. Therefore, ABCD is a parallelogram.

Problem 4 and 5: Due to ambiguous diagrams and possible typos, I recommend checking the diagram carefully. For Problem 4, if you can identify point G, use triangle congruence criteria with the given congruences. Similarly for Problem 5.

But since the user asked to solve accurately, and for Problem 1,2,3 are clear, I'll box those.

Final Answer for Problem 1 MC: 3

For proofs, they are written above.

But the instruction is to provide final answer after reasoning.

Since the main multiple choice is Problem 1, and others are proofs, I'll state:

Final Answer for the multiple choice question in Problem 1: 3

For the proofs, they are completed in the reasoning.

To comply with format:

Final Answer:
3