Proof of Similar Triangles 1 | Download Free PDF | Euclid ... - Free Printable

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

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Problem 1: Flow Proof for △JKL ~ △PLM

Given: ∠J ≅ ∠P
We need to prove: △JKL ~ △PLM using a flow proof.

Step-by-step reasoning:

- Look at the diagram: triangles JKL and PLM share point L, and lines JM and KP cross at L. That means ∠JLK and ∠PLM are vertical angles → so they are congruent.
- We’re given that ∠J ≅ ∠P.
- If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar (AA Similarity Postulate).
- So we have:
- ∠J ≅ ∠P (given)
- ∠JLK ≅ ∠PLM (vertical angles)
→ Therefore, △JKL ~ △PLM by AA~

Flow proof structure (written as steps with arrows):

∠J ≅ ∠P → Given
∠JLK ≅ ∠PLM → Vertical Angles Theorem
△JKL ~ △PLM → AA Similarity Postulate

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Problem 2: Two Column Proof for △YXZ ~ △BAC

Given: △XYZ and △ABC are right triangles; XY/AB = YZ/BC
Prove: △YXZ ~ △BAC

Note: The order matters in similarity — △YXZ corresponds to △BAC, meaning:
- Y ↔ B
- X ↔ A
- Z ↔ C

So we need to match sides and angles accordingly.

Step-by-step:

- Both triangles are right triangles → so they each have a right angle.
- In △XYZ, right angle is at Y (from diagram: square at Y)
- In △ABC, right angle is at B (square at B)
→ So ∠Y ≅ ∠B (both are right angles → 90°)

Also given: XY / AB = YZ / BC

Look at the sides around the right angles:

In △YXZ:
- Legs are XY and YZ (since right angle at Y)

In △BAC:
- Legs are BA and BC? Wait — let’s check correspondence.

Wait! The triangles are named △YXZ and △BAC.

So vertex Y corresponds to B, X to A, Z to C.

That means side YX corresponds to BA, and YZ corresponds to BC.

Given: XY / AB = YZ / BC → same as YX / BA = YZ / BC

And included angle between those sides: ∠Y and ∠B — both right angles → congruent.

So we have two sides proportional and the included angle congruent → SAS Similarity!

Two column proof:

| Statements | Reasons |
|-------------------------------|----------------------------------|
| 1. △XYZ and △ABC are right triangles | Given |
| 2. ∠Y and ∠B are right angles | Definition of right triangle |
| 3. ∠Y ≅ ∠B | All right angles are congruent |
| 4. XY / AB = YZ / BC | Given |
| 5. △YXZ ~ △BAC | SAS Similarity (steps 3 & 4) |

Note: Step 5 uses SAS~ because we have two sides proportional and the included angle congruent.

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Problem 3: Two Column Proof for PQ/SQ = TQ/RQ

Given: PT || SR
Prove: PQ / SQ = TQ / RQ

Diagram shows two triangles sharing point Q: △PQT and △SQR, with PT parallel to SR.

This looks like an “X” shape with parallel lines — classic setup for similar triangles via alternate interior angles.

Step-by-step:

- Since PT || SR, and line PS crosses them → alternate interior angles are congruent.
- Specifically, ∠TPQ ≅ ∠RSQ (alternate interior angles)
- Also, ∠PTQ ≅ ∠SRQ (another pair of alternate interior angles)
- Also, ∠PQT and ∠SQR are vertical angles → congruent.
- So actually, we can use AA~ to show △PQT ~ △SQR.

Once triangles are similar, corresponding sides are proportional.

Corresponding vertices:
- P ↔ S
- Q ↔ Q (common vertex, but actually it's shared angle)
Wait — better to map based on angles.

From above:
- ∠TPQ ∠RSQ → so P corresponds to S
- ∠PTQ ≅ ∠SRQ → so T corresponds to R
- Then Q corresponds to Q

So △PQT ~ △SQR

Therefore, sides:
PQ corresponds to SQ
TQ corresponds to RQ
PT corresponds to SR

So proportion: PQ / SQ = TQ / RQ = PT / SR

Which is exactly what we need to prove: PQ / SQ = TQ / RQ

Two column proof:

| Statements | Reasons |
|--------------------------------|-------------------------------------|
| 1. PT || SR | Given |
| 2. ∠TPQ ≅ ∠RSQ | Alternate Interior Angles Theorem |
| 3. ∠PTQ ≅ ∠SRQ | Alternate Interior Angles Theorem |
| 4. △PQT ~ △SQR | AA Similarity Postulate |
| 5. PQ / SQ = TQ / RQ | Corresponding parts of similar triangles are proportional |

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Final Answer:

Problem 1 Flow Proof:
∠J ≅ ∠P → Given
∠JLK ≅ ∠PLM → Vertical Angles
△JKL ~ △PLM → AA Similarity

Problem 2 Two Column Proof:

| Statements | Reasons |
|-------------------------------|----------------------------------|
| 1. △XYZ and △ABC are right triangles | Given |
| 2. ∠Y and B are right angles | Definition of right triangle |
| 3. ∠Y ≅ ∠B | All right angles are congruent |
| 4. XY / AB = YZ / BC | Given |
| 5. △YXZ ~ △BAC | SAS Similarity |

Problem 3 Two Column Proof:

| Statements | Reasons |
|--------------------------------|-------------------------------------|
| 1. PT || SR | Given |
| 2. ∠TPQ ≅ ∠RSQ | Alternate Interior Angles Theorem |
| 3. ∠PTQ ∠SRQ | Alternate Interior Angles Theorem |
| 4. △PQT ~ △SQR | AA Similarity Postulate |
| 5. PQ / SQ = TQ / RQ | Corresponding parts of similar triangles are proportional |