Worksheet titled "Congruent Triangles Worksheet" from Math Monks, featuring six problems that require students to prove triangle congruence using various theorems and given information such as side lengths, angles, and geometric figures.

Let’s go through each problem one by one, step by step.

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Problem 1:

We are given two triangles:

Triangle ABC:
- AB = 2 yd
- BC = 3.5 yd
- AC = 4.5 yd

Triangle PQR:
- PQ = 2 yd
- QR = 4.5 yd
- PR = 3.5 yd

Wait — let’s match the sides carefully.

In triangle ABC: sides are 2, 3.5, 4.5
In triangle PQR: sides are 2, 4.5, 3.5 → same three lengths!

So all three sides of triangle ABC match the three sides of triangle PQR (just in different order).

That means we can use SSS (Side-Side-Side) congruence theorem.

But we must make sure the correspondence is correct.

Let’s map:

AB = 2 → matches PQ = 2
BC = 3.5 → matches PR = 3.5? Wait, no — BC is between B and C, which are connected to A.

Actually, better to list sides with vertices:

Triangle ABC:
AB = 2, BC = 3.5, AC = 4.5

Triangle PQR:
PQ = 2, QR = 4.5, PR = 3.5

So if we match:

A ↔ P (since AB = PQ = 2)
B ↔ Q
Then C should correspond to R? But then AC = 4.5 should equal PR = 3.5? No, that doesn’t match.

Wait — maybe A ↔ P, B ↔ Q, but then C ↔ R? Then AC should be PR? But AC=4.5, PR=3.5 — not equal.

Alternatively, try matching:

AB = 2 → PQ = 2 → so A↔P, B↔Q
BC = 3.5 → QR = 4.5? No.

Wait — perhaps the labeling is not in order. Let’s just compare sets of side lengths.

Both triangles have sides: 2, 3.5, 4.5 → so they are congruent by SSS.

To write the congruence statement correctly, we need to match corresponding vertices.

Look at triangle ABC: shortest side is AB=2, middle is BC=3.5, longest is AC=4.5

Triangle PQR: PQ=2 (shortest), PR=3.5 (middle), QR=4.5 (longest)

So:

AB (2) ↔ PQ (2) → A↔P, B↔Q
BC (3.5) ↔ PR (3.5)? But BC connects B-C, PR connects P-R → so if B↔P? That conflicts.

Better approach: since all three sides are equal, the triangles are congruent. The correspondence might be:

A ↔ Q? Let’s check distances.

Actually, let’s assign based on side lengths:

In triangle ABC:
- Side opposite A is BC = 3.5
- Side opposite B is AC = 4.5
- Side opposite C is AB = 2

In triangle PQR:
- Side opposite P is QR = 4.5
- Side opposite Q is PR = 3.5
- Side opposite R is PQ = 2

So:

Angle A (opposite 3.5) ↔ angle Q (opposite 3.5)
Angle B (opposite 4.5) ↔ angle P (opposite 4.5)
Angle C (opposite 2) ↔ angle R (opposite 2)

Therefore, correspondence: A↔Q, B↔P, C↔R

So triangle ABC ≅ triangle QPR

But usually we write in order: so ΔABC ≅ ΔQPR

But the question says “state the theorem” — it’s SSS.

Answer for #1: Yes, congruent by SSS. Correspondence: A→Q, B→P, C→R → so ΔABC ≅ ΔQPR

But let’s double-check the diagram description — actually, looking back, in the image, triangle ABC has AB=2, BC=3.5, AC=4.5; triangle PQR has PQ=2, QR=4.5, PR=3.5.

So if we take:

AB = PQ = 2
AC = QR = 4.5
BC = PR = 3.5

Then correspondence: A↔P, B↔Q, C↔R? Then AC should be PR? No.

If A↔P, B↔Q, then C↔R → then AC corresponds to PR? But AC=4.5, PR=3.5 — not equal.

Wait — perhaps A↔P, C↔Q, B↔R?

Try:

Set A↔P (both have side 2 from them)
From A: AB=2, AC=4.5
From P: PQ=2, PR=3.5 — not matching.

This is messy. Since all three sides are equal, regardless of labeling, the triangles are congruent by SSS. For the purpose of this worksheet, we can say:

Yes, the triangles are congruent by SSS theorem.

Corresponding vertices: Let’s match the sides:

AB = 2, PQ = 2 → so A-P, B-Q
Then BC = 3.5, and in other triangle, PR = 3.5 — so if B-Q, then C must correspond to R? But then AC should be QR? AC=4.5, QR=4.5 — yes!

So:

A ↔ P
B ↔ Q
C ↔ R

Check:

AB = PQ = 2
BC = QR? BC=3.5, QR=4.5 — no! Wait, QR is 4.5, but BC is 3.5.

I think I made a mistake earlier.

Looking again at the problem as written:

Triangle ABC: AB=2, BC=3.5, AC=4.5
Triangle PQR: PQ=2, QR=4.5, PR=3.5

So sides:

ABC: AB=2, BC=3.5, CA=4.5
PQR: PQ=2, QR=4.5, RP=3.5

Now, if we match:

AB = PQ = 2
BC = RP = 3.5
CA = QR = 4.5

So correspondence: A↔P, B↔Q, C↔R? Then BC should be QR, but BC=3.5, QR=4.5 — no.

If A↔P, B↔R, C↔Q?

Then AB = PR? AB=2, PR=3.5 — no.

Let’s do it properly.

List the sides with endpoints:

For triangle ABC:
- Side between A and B: 2
- Side between B and C: 3.5
- Side between C and A: 4.5

For triangle PQR:
- Side between P and Q: 2
- Side between Q and R: 4.5
- Side between R and P: 3.5

So, the side of length 2 is between A-B and P-Q → so A-B corresponds to P-Q

The side of length 3.5 is between B-C and R-P → so B-C corresponds to R-P

The side of length 4.5 is between C-A and Q-R → so C-A corresponds to Q-R

So, from A-B ↔ P-Q → A↔P, B↔Q

From B-C ↔ R-P → B↔R, C↔P — conflict with above.

From C-A ↔ Q-R → C↔Q, A↔R — another conflict.

This suggests that the correspondence is not direct. Perhaps the triangles are congruent but the vertex labels are not in corresponding order.

Since all three sides are equal, they are congruent by SSS. We can say:

ΔABC ≅ ΔPQR is not correct because sides don't match in that order.

But ΔABC ≅ ΔQRP or something.

Let’s define the correspondence where:

Vertex A corresponds to vertex Q (because both are between sides of length 2 and 4.5? In ABC, A is between AB=2 and AC=4.5. In PQR, Q is between PQ=2 and QR=4.5 — yes!

Similarly, B is between AB=2 and BC=3.5 — in PQR, P is between PQ=2 and PR=3.5 — so B↔P

C is between BC=3.5 and AC=4.5 — in PQR, R is between PR=3.5 and QR=4.5 — so C↔R

So correspondence: A↔Q, B↔P, C↔R

Thus, ΔABC ≅ ΔQPR

And the theorem is SSS.

So for problem 1: Yes, congruent by SSS. ΔABC ≅ ΔQPR

But typically, we might write it as ΔABC ≅ ΔQPR, but let's confirm the side lengths under this correspondence:

AB = 2, QP = 2 — good
BC = 3.5, PR = 3.5 — good
AC = 4.5, QR = 4.5 — good

Perfect.

So answer for #1: Yes, congruent by SSS. ΔABC ≅ ΔQPR

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Problem 2:

Prove ΔAEB ≅ ΔAEC

Given figure: triangle ABC with AE perpendicular to BC, and AB = AC = 7 in. Also, BE and EC are parts of BC, and there's a right angle at E.

So, we have:

- AB = AC = 7 in (given)
- AE is common to both triangles AEB and AEC
- Angle at E is 90 degrees for both, so ∠AEB = ∠AEC = 90°

So, in triangles AEB and AEC:

- AB = AC (hypotenuse)
- AE = AE (common leg)
- Right angles at E

So, by HL (Hypotenuse-Leg) congruence for right triangles, ΔAEB ≅ ΔAEC

HL is a special case for right triangles.

We could also use SAS: since AE is common, angles at E are both 90°, and if we knew BE=EC, but we don't necessarily know that yet — but from congruence, we can prove it.

But here, since we have hypotenuse and one leg equal, HL applies.

So proof: In ΔAEB and ΔAEC,

- AB = AC (given)
- AE = AE (reflexive property)
- ∠AEB = ∠AEC = 90° (given, since AE ⊥ BC)

Therefore, by HL congruence, ΔAEB ≅ ΔAEC.

Note: Some curricula accept HL, others might want to use Pythagoras to show BE=EC first, but HL is standard.

So answer for #2: Proven by HL congruence.

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Problem 3:

Given: AB ≅ EF, BC ≅ DF

Show that ΔABD ≅ ΔEFC

Figure: Points B, C, D, F on a line, with AB perpendicular to BF at B, EF perpendicular to BF at F. Also, BC = CD? Wait, markings: on BD, there are tick marks: from B to C has one tick, C to D has one tick? Looking at description: "in the above figure" — but since no image, from text: "AB ≅ EF and BC ≅ DF"

Also, from typical such problems, likely B-C-D-F colinear, with AB ⊥ BD at B, EF ⊥ FD at F, and BC = DF, and perhaps CD is common or something.

The triangles are ΔABD and ΔEFC.

Points: A-B-D and E-F-C.

Given AB = EF, BC = DF.

But BC and DF are not directly sides of the triangles.

Perhaps we need to find more.

Assume the figure has points B, C, D, F on a straight line, with B-C-D-F, and AB perpendicular at B, EF perpendicular at F.

Markings: probably BC has one tick, CD has one tick? Or what.

The problem says: "in the above figure AB ≅ EF and BC ≅ DF"

And we need to show ΔABD ≅ ΔEFC.

First, note that ΔABD has points A,B,D; ΔEFC has E,F,C.

Sides: for ΔABD: AB, BD, DA

For ΔEFC: EF, FC, CE

Given AB = EF.

Now, BD = BC + CD

FC = FD + DC? If points are B,C,D,F in order, then FC = FD + DC only if D is between F and C, which may not be.

Probably the order is B, C, D, F on a line, with C between B and D, D between C and F? Unclear.

From the congruence to prove: ΔABD and ΔEFC.

Perhaps C and D are positioned such that BD and FC are related.

Another thought: perhaps BC = DF, and CD is common, so BD = BC + CD, FC = FD + DC = DF + CD = BC + CD = BD, since BC=DF.

Yes! If points are colinear in order B, C, D, F, then:

BD = BC + CD

FC = FD + DC = DF + CD (since FD=DF)

But given BC = DF, so BD = BC + CD = DF + CD = FC

So BD = FC

Also, AB = EF (given)

And angles at B and F are both 90 degrees, since AB ⊥ BD and EF ⊥ FC (assuming from figure).

So in ΔABD and ΔEFC:

- AB = EF (given)
- BD = FC (as shown above)
- ∠ABD = ∠EFC = 90° (right angles)

Therefore, by SAS congruence, ΔABD ≅ ΔEFC.

SAS: two sides and included angle.

Here, AB and BD with included angle at B; EF and FC with included angle at F.

Yes.

So answer for #3: By SAS, since AB=EF, BD=FC (because BC=DF and CD common), and right angles at B and F.

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Problem 4:

Given ΔDAB ≅ ΔBCD

Find x.

Figure: quadrilateral ABCD, with diagonal DB. Angles given: at A, angle DAB = 2x² + 7; at C, angle BCD = 57°.

Since ΔDAB ≅ ΔBCD, their corresponding angles are equal.

Need to find correspondence.

ΔDAB and ΔBCD.

Vertices: D,A,B and B,C,D.

Probably correspondence D↔B, A↔C, B↔D? Let's see.

Since it's ΔDAB ≅ ΔBCD, likely D↔B, A↔C, B↔D.

So angle at A corresponds to angle at C.

Angle DAB corresponds to angle BCD.

Because in ΔDAB, angle at A is between D and B; in ΔBCD, angle at C is between B and D.

So yes, ∠DAB = ∠BCD

Given ∠DAB = 2x² + 7, ∠BCD = 57°

So 2x² + 7 = 57

Solve:

2x² = 57 - 7 = 50

x² = 25

x = 5 or x = -5

Since it's an angle measure, probably positive, so x=5.

Check: 2*(25) +7 = 50+7=57, yes.

So x=5.

Answer for #4: x = 5

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Problem 5:

Given ΔABC ≅ ΔXYZ under ASA.

Find x and y.

Triangles:

ΔABC: angle B=60°, angle C=30°, side BC=6 cm

ΔXYZ: angle Y=60°, angle Z=x, side YZ=6 cm, and angle at X is y? Wait, labeled: in ΔXYZ, at Y is 60°, at Z is x, and side YZ=6 cm, and angle at X is y? The diagram shows: point X, Y, Z; angle at Y is 60°, angle at Z is marked as x, side YZ=6 cm, and angle at X is marked as y.

Given congruence under ASA.

ASA means two angles and the included side.

In ΔABC: angles at B and C are given: 60° and 30°, and side BC=6 cm, which is between B and C.

So included side between the two angles.

Similarly, in ΔXYZ, side YZ=6 cm, which is between Y and Z.

Angles at Y and Z are given: at Y is 60°, at Z is x.

Since congruent under ASA, and correspondence is ΔABC ≅ ΔXYZ, so A↔X, B↔Y, C↔Z.

So angle B = angle Y = 60° — matches.

Angle C = angle Z = 30° — so x = 30°

Side BC = side YZ = 6 cm — matches.

Now, angle at A: in ΔABC, sum of angles is 180°, so angle A = 180 - 60 - 30 = 90°

Similarly, in ΔXYZ, angle at X = y = 180 - 60 - x = 180 - 60 - 30 = 90°

So y=90°

The problem says "find the value of x and y", and in the diagram, x is at Z, y is at X.

So x=30, y=90.

But let's confirm ASA: in ΔABC, angles at B and C with included side BC.

In ΔXYZ, angles at Y and Z with included side YZ.

Yes, and since B↔Y, C↔Z, it matches.

So x=30, y=90.

Answer for #5: x = 30, y = 90

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Problem 6:

Complete the congruence statement.

Part a): Triangle ABC and triangle ADC? Diagram: triangle ABD with C on BD, AC perpendicular to BD, and AB=AD (tick marks), BC=CD (tick marks), and AC common.

So, triangles ABC and ADC.

Given: AB = AD (marked), BC = CD (marked), AC = AC (common), and angle at C is 90° for both.

So, by SSS or SAS or HL.

Specifically, since right-angled at C, and legs BC=CD, hypotenuse AB=AD, so HL.

Or SAS: BC=CD, angle C=90°, AC=AC.

So ΔABC ≅ ΔADC

Correspondence: A↔A, B↔D, C↔C? But C is common.

Vertices: A,B,C and A,D,C.

So ΔABC ≅ ΔADC

With A↔A, B↔D, C↔C.

Yes.

So ΔABC ≅ ΔADC

Part b): Two triangles: PQT and QRS? Diagram: points P,Q,R on a line, with T and S below. Markings: PT has one tick, QT has two ticks, QS has two ticks, RS has one tick. Angles: at P and R are marked equal (single arc), at Q for both triangles are marked equal (double arc?).

Triangles are ΔPQT and ΔQRS? Or ΔPQT and ΔSQR?

The congruence to complete is ΔQRS ≅ _____

From diagram: likely ΔPQT and ΔSQR or something.

Points: P, Q, R colinear, T and S on the other side.

Triangle PQT and triangle SQR? Or QRS.

Given: PT = RS (one tick each), QT = QS (two ticks each), and angle at Q for both triangles: angle PQT and angle SQR are vertical angles? Or adjacent.

Actually, since P,Q,R colinear, and T and S on opposite sides, angle at Q for triangle PQT is angle PQT, for triangle QRS is angle RQS.

But angle PQT and angle RQS are vertically opposite if T and S are on opposite sides, but probably not.

Markings: angle at P and angle at R are equal (both have single arc), angle at Q for triangle PQT has double arc, angle at Q for triangle QRS has double arc? In the description, it says "angles at P and R are marked", and "at Q" for both.

Assuming that angle TPQ = angle SRQ (given by arcs), and angle PQT = angle RQS (vertically opposite or given), and sides.

Specifically, PT = RS (one tick), QT = QS (two ticks), and angle at Q: if angle PQT and angle RQS are vertically opposite, they are equal.

In the figure, since P-Q-R straight line, and T and S on the same side or opposite? Typically, if T and S are on opposite sides of PR, then angle PQT and angle RQS are vertically opposite, hence equal.

So, in ΔPQT and ΔRQS:

- PT = RS (given)
- QT = QS (given)
- angle PQT = angle RQS (vertically opposite angles)

So by SAS, ΔPQT ≅ ΔRQS

But the question asks for ΔQRS ≅ _____

ΔQRS is the same as ΔRQS.

So ΔQRS ≅ ΔPQT

Correspondence: Q↔P, R↔Q, S↔T? Let's see.

In ΔQRS and ΔPQT:

Side QR corresponds to? Not directly.

From SAS: in ΔPQT and ΔRQS:

PQ corresponds to RQ? Not necessarily.

With PT=RS, QT=QS, angle at Q equal.

So the correspondence is P↔R, Q↔Q, T↔S? But then PT corresponds to RS, which is good, QT to QS, good, angle at Q same.

But angle at Q is between PQ and QT in first triangle, between RQ and QS in second.

If P↔R, Q↔Q, T↔S, then side PQ corresponds to RQ, but we don't know if PQ=RQ.

In the diagram, there might be no marking for PQ and RQ, so probably not equal.

Perhaps correspondence is P↔S, Q↔Q, T↔R? Messy.

From the SAS: the two sides and included angle.

In ΔPQT, sides PT and QT with included angle at T? No, angle at Q is between PQ and QT.

Standard SAS: two sides and the included angle.

Here, for ΔPQT, we have sides QT and PT, but the angle between them is at T, not at Q.

I think I confused.

In triangle PQT, the sides from Q are QP and QT, and the angle between them is angle PQT.

Similarly, in triangle RQS, sides from Q are QR and QS, angle between them is angle RQS.

If angle PQT = angle RQS, and if QP = QR and QT = QS, then SAS.

But we have QT = QS, but not necessarily QP = QR.

In the diagram, there are no markings on QP and QR, so probably not equal.

But we have PT = RS, which is not adjacent to angle at Q.

Perhaps it's SSA, which is not valid.

Another possibility: the triangles are ΔPQT and ΔSQR.

Let me denote.

Assume points: P--Q--R on a line.

T is below, connected to P and Q.

S is below, connected to Q and R.

So triangle PQT and triangle SQR.

Given: PT = RS (one tick), QT = QS (two ticks), and angle at P = angle at R (arcs), and angle at Q for both: angle PQT and angle SQR.

Angle at Q for triangle PQT is angle between PQ and QT.

For triangle SQR, angle at Q is between SQ and QR.

These are not the same angle.

However, if we consider triangle PQT and triangle RQS, then angle at Q is the same vertex.

Perhaps the congruence is ΔPQT ≅ ΔRQS by SAS if we have the sides.

But we have PT = RS, QT = QS, and if angle at T equals angle at S, but not given.

Notice that in the diagram, there is also angle at T and S marked? The problem says "angles at P and R are marked", and "at Q" , but not at T and S.

Perhaps it's ASA or AAS.

Another idea: since QT = QS, and if angle at Q is common or something.

Let's look at the angles given.

In triangle PQT, angles: at P is marked, at Q is marked.

In triangle QRS, angles: at R is marked, at Q is marked.

And since P-Q-R straight, the angles at Q for the two triangles are adjacent or vertical.

Typically in such diagrams, angle PQT and angle RQS are vertically opposite if T and S are on opposite sides, but here probably on the same side, so they are adjacent.

Perhaps angle PQT and angle SQR are the same if S and T are symmetric, but not specified.

I recall that in many textbooks, this is a standard problem where ΔPQT ≅ ΔSQR by SAS or something.

Let's assume the correspondence.

Suppose ΔPQT ≅ ΔSQR.

Then P↔S, Q↔Q, T↔R.

Then PT corresponds to SR, which is given PT=RS, so SR=RS, good.

QT corresponds to QR? But QT=QS, not necessarily QR.

Not good.

ΔPQT ≅ ΔRQS.

P↔R, Q↔Q, T↔S.

Then PT corresponds to RS, good.

QT corresponds to QS, good.

Angle at Q: angle PQT corresponds to angle RQS.

If angle PQT = angle RQS, then SAS holds.

And in the diagram, since P-Q-R is straight, and if T and S are on the same side, then angle PQT and angle RQS are not necessarily equal, but if the figure is symmetric, or from the markings, the angles at Q are marked with the same number of arcs, so likely angle PQT = angle RQS.

In the problem description, it says "angles at P and R are marked" with single arc, and "at Q" for both triangles are marked with double arc, so probably angle at Q in both triangles are equal.

So yes, angle PQT = angle RQS.

And sides QT = QS, PT = RS.

But for SAS, we need the included angle between the two sides.

In triangle PQT, sides QT and PT, the included angle is at T, not at Q.

Mistake.

To use SAS with angle at Q, we need the two sides forming that angle, which are QP and QT for triangle PQT.

But we don't have QP = anything.

Unless we have another pair.

Perhaps it's AAS.

In triangle PQT and triangle RQS:

- angle at P = angle at R (given)
- angle at Q = angle at Q (given, and if they are the same measure)
- side QT = QS (given)

But side QT is not between the angles; in AAS, we need a side not between the angles.

Specifically, for AAS, if two angles and a non-included side are equal.

Here, in ΔPQT and ΔRQS:

- angle P = angle R
- angle Q = angle Q (assume equal)
- side QT = QS

Now, side QT is opposite to angle P in triangle PQT? Let's see.

In triangle PQT, side opposite to angle P is QT.

In triangle RQS, side opposite to angle R is QS.

Since angle P = angle R, and QT = QS, and angle Q = angle Q, then by AAS, the triangles are congruent.

AAS congruence: if two angles and a non-included side of one triangle are equal to those of another, then congruent.

Here, for ΔPQT and ΔRQS:

- angle at P = angle at R
- angle at Q = angle at Q
- side QT = side QS, and QT is opposite to angle P, QS is opposite to angle R, and since angle P = angle R, it works.

More precisely, the side is corresponding.

Since angle P = angle R, and angle Q = angle Q, then the third angles are equal, and with one side equal, it should be fine.

Specifically, side QT is between angles Q and T, but in AAS, as long as the side is corresponding.

Standardly, if two angles and any side are equal, it determines the triangle.

So yes, by AAS, ΔPQT ≅ ΔRQS.

Therefore, for the congruence statement, ΔQRS is the same as ΔRQS, so ΔQRS ≅ ΔPQT

With correspondence Q↔P, R↔Q, S↔T? From above, R↔P, Q↔Q, S↔T, so ΔRQS ≅ ΔPQT, so ΔQRS ≅ ΔPQT

Yes.

So for part b: ΔQRS ≅ ΔPQT

Now, to write the answers.

Summary:

1. Yes, congruent by SSS. ΔABC ≅ ΔQPR

2. Proven by HL congruence.

3. By SAS, since AB=EF, BD=FC (as BC=DF and CD common), and right angles at B and F.

4. x = 5

5. x = 30, y = 90

6. a) ΔABC ≅ ΔADC
b) ΔQRS ≅ ΔPQT

Now, for the final answer, since the user didn't specify which problem, but the instruction is to solve the problem, and there are multiple, I'll provide all.

But in the response format, it says "Final Answer" after explanations, but since ExplainAnswer is false, I shouldn't explain, but the critical rule says to work through step by step before final answer.

The user options are all false, so according to rules, I should omit Explanation, TeachTopic, GeneratePractice sections.

But the critical rule says: "First work through the problem step by step, check your calculations carefully, and ONLY THEN provide the final answer."

And "Complete ALL reasoning and explanations BEFORE writing the Final Answer."

But if ExplainAnswer is false, I shouldn't include explanation section.

There's a conflict.

Looking back at the response format rules:

"If ExplainAnswer is true: [explanation section]"

Similarly for others.

And "If an option is false, completely omit that section."

But the critical rule says to work through step by step before final answer.

Perhaps for this case, since all options are false, I should still do the reasoning internally, but only output the final answer.

But the critical rule emphasizes to verify before final answer.

To be safe, I'll provide the final answers for each problem.

Since the worksheet has 6 problems, I'll list the answers.

Final Answer:

1. Yes, congruent by SSS; \triangle ABC \cong \triangle QPR

2. \triangle AEB \cong \triangle AEC by HL congruence

3. \triangle ABD \cong \triangle EFC by SAS congruence

4. x = 5

5. x = 30, y = 90

6. a) \triangle ABC \cong \triangle ADC
b) \triangle QRS \cong \triangle PQT

For problem 1, I had ΔABC ≅ ΔQPR, but let's confirm the correspondence.

Earlier I said A↔Q, B↔P, C↔R, so ΔABC ≅ ΔQPR.

Sometimes written as ΔABC ≅ ΔQPR.

For problem 6b, ΔQRS ≅ ΔPQT.

I think that's correct.

So boxing the final answers.