Geometric Proof and Angles Worksheet for 10th Grade | Lesson Planet - Free Printable

Let’s go through each proof step by step. I’ll explain what’s happening in simple terms so you can follow along easily.

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PRACTICE #1 (Left Side – SSS Proof)

We are given:
- AD ≅ CD
- B is the midpoint of AC → that means AB = BC
We need to prove: ΔABD ≅ ΔCBD

Step-by-step:

1. AD ≅ CD — This was given to us. So we write “Given” as the reason.
2. B is the midpoint of AC — Also given. That tells us something important about segments AB and BC.
3. AB ≅ CB — Because B is the midpoint, it splits AC into two equal parts. So AB and CB are congruent. Reason: Definition of midpoint.
4. BD ≅ BD — This is the same segment in both triangles! It’s shared. So it’s congruent to itself. Reason: Reflexive Property (anything is congruent to itself).
5. ΔABD ≅ ΔCBD — Now we have all three sides of one triangle congruent to all three sides of the other triangle:
- AD ≅ CD (step 1)
- AB ≅ CB (step 3)
- BD ≅ BD (step 4)
So we use SSS (Side-Side-Side) Congruence Postulate. Reason: SSS (using steps 1, 3, 4).

✔ Done! The triangles are congruent by SSS.

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PRACTICE #2 (Left Side – Another SSS Proof)

Given:
- PR ≅ NT
- NO ≅ SR
- O is halfway from N to P → so NP is split in half at O? Wait — actually, looking at the diagram and statement, it says “O is ½ of the way from N to P” and “S is ½ of the way from R to T”. That probably means NO = OP and RS = ST? But wait — let’s read carefully.

Actually, the key is:
“O is ½ of the way from N to P” → so NO = OP → meaning NP = 2·NO
Similarly, “S is ½ of the way from R to T” → so RS = ST → RT = 2·RS

But we’re told NO ≅ SR → so if NO = SR, then 2·NO = 2·SR → so NP = RT → so NP ≅ RT

Then we also have reflexive property for NR ≅ NR (same side)

So let’s walk through the proof:

1. PR ≅ NT — Given
2. NO ≅ SR — Given
3. NP ≅ RT — How? Since O is midpoint of NP → NP = 2·NO; S is midpoint of RT → RT = 2·SR. And since NO ≅ SR, then 2·NO ≅ 2·SR → so NP ≅ RT. Reason: Multiplication Property (if two things are equal, multiplying both by 2 keeps them equal).
4. NR ≅ NR — Same segment, so reflexive property.
5. ΔNRT ≅ ΔRNP — Now check sides:
- NR ≅ NR (step 4)
- RT ≅ NP (step 3)
- NT ≅ PR (step 1)
So all three sides match → SSS congruence. Reason: SSS (steps 1, 3, 4)

Wait — hold on! The triangles are named ΔNRT and ΔRNP. Let’s make sure the correspondence is right.

In ΔNRT: sides are NR, RT, TN
In ΔRNP: sides are RN, NP, PR

We have:
- NR ≅ RN (same thing)
- RT ≅ NP (from step 3)
- TN ≅ PR (given as NT ≅ PR — same thing)

Yes, so SSS works.

✔ Done!

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PRACTICE #1 (Right Side – SAS & HL)

Given:
- ∠1 ≅ ∠2
- M is midpoint of BC → so BM ≅ MC
- BE ≅ CE
Prove: ΔEMB ≅ ΔEMC

Steps:

1. BE ≅ CE — Given
2. ∠1 ≅ ∠2 — Given
3. ∠1 is supp. to ∠3; ∠2 is supp. to ∠4 — From diagram, angles 1 and 3 form a straight line → they are supplementary. Same with 2 and 4. Reason: If two angles form a straight angle, they are supplementary.
4. ∠3 ≅ ∠4 — Why? Because ∠1 ≅ ∠2 (given), and supplements of congruent angles are congruent. So if ∠1 + ∠3 = 180° and ∠2 + ∠4 = 180°, and ∠1 = ∠2, then ∠3 must equal ∠4. Reason: Congruent Supplements Theorem.
5. M is midpoint of BC — Given
6. MB ≅ MC — By definition of midpoint.
7. ΔEMB ≅ ΔEMC — Now look:
- We have BE ≅ CE (side)
- ∠3 ≅ ∠4 (angle between those sides?) Wait — let’s see the diagram.

Actually, in triangle EMB and EMC:

- Side: BE ≅ CE (given)
- Angle: ∠3 ≅ ∠4 (just proved) — these are the angles at B and C?
Wait — no. Looking at the diagram: points A-B-M-C-D on a line. E is above. So triangle EMB has vertices E, M, B. Triangle EMC has E, M, C.

Angle at B in triangle EMB is ∠EBM — which is ∠3? Yes, because ∠1 and ∠3 are adjacent on the line. Similarly, ∠2 and ∠4 are adjacent.

Actually, ∠3 is angle EBM, and ∠4 is angle ECM.

And we have:
- BE ≅ CE (side)
- ∠3 ≅ ∠4 (angle)
- BM ≅ CM (side)

So that’s Side-Angle-Side: two sides and the included angle.

Wait — is the angle included? In triangle EMB: sides EB and MB with included angle at B (∠3). In triangle EMC: sides EC and MC with included angle at C (∠4). Yes! So SAS applies.

Reason: SAS (steps 1, 4, 6)

✔ Done!

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PRACTICE #2 (Right Side – HL Proof)

Given:
- BC ⊥ AC → so angle at C is right angle
- BD ⊥ AD → so angle at D is right angle
- AC ≅ AD
Prove: ΔACB ≅ ΔADB

Steps:

1. BC ⊥ AC, BD ⊥ AD — Given
2. ∠ACB is a rt. ∠, ∠BDA is a rt. ∠ — Perpendicular lines form right angles. Reason: Definition of perpendicular lines.
3. ΔACB and ΔADB are right triangles — Because they each have a right angle. Reason: Definition of right triangles.
4. AC ≅ AD — Given
5. AB ≅ AB — Shared hypotenuse! Both triangles share side AB. Reason: Reflexive Property.
6. ΔACB ≅ ΔADB — Now, both are right triangles. They have:
- One leg congruent: AC ≅ AD (step 4)
- Hypotenuse congruent: AB ≅ AB (step 5)
So we use HL (Hypotenuse-Leg) Congruence Theorem for right triangles.

✔ Done!

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Final Answer:
All proofs are completed correctly using the appropriate congruence postulates:
- Left Practice #1: SSS
- Left Practice #2: SSS
- Right Practice #1: SAS
- Right Practice #2: HL

Each step follows logically from the given information and geometric definitions/properties.