# Congruent Triangle Problems by Angle-Side-Angle and Side-Side-Side

#### Angle-Side-Angle(ASA)

#### Angle-Angle-Side(AAS)

#### Side-Side-Side(SSS)

## Problems

**Problem 1**

Given ABC - an isosceles triangle.

AM and BN are angle bisectors.

Prove that AM ≅ BN.

**Solution:**

Triangles AMB and BNA are congruent (by Angle-Side-Angle) because:

1. ∠CAB ≅ ∠CBA

2. AB – is shared.

3. ∠MAB ≅ ∠NBA = $\frac{1}{2}$∠CAB

The segments AM and BN are corresponding sides in these congruent triangles and therefore AM ≅ BN.

**Problem 2**

Given:

ABC is a triangle.

CM is a median.

AA_{1} ⊥ CM and BB_{1} ⊥ CM.

Prove that AA_{1} ≅ BB_{1}

**Solution:**

Triangles AA_{1}M and
BB_{1}M are congruent (by Angle-Side-Angle) because

∠BB_{1}M ≅ ∠AA_{1}M = 90°,

∠AMA_{1} ≅ ∠BMA_{1},

AM ≅ BM

Therefore, AA_{1} ≅ BB_{1}.

**Problem 3**

Prove that the perpendiculars built from any point of the bisector of an angle towards its shoulders makes equal segments.

**Solution:**

Let M be an arbitrary point on the bisector $\vec{OL}$, $\overline{MP} \perp \vec{OA}$ and $\overline{MQ} \perp \vec{OB}$

To prove that OP ≅ OQ is enough to prove that △OPM ≅
△OQM.

1) ∠QOM ≅ ∠POM (OL is a bisector),

2) ∠OQM ≅ ∠OPM = 90°

3) OM is a shared side.

Therefore, the both triangles OPM and OQM are congruent by Angle-Angle-Side.

**Problem 4**

Prove that if CP is an altitude and a bisector then the triangle ABC is isosceles.

**Solution:**

If AC ≅ BC then △ABC is isosceles.

Let's look at the triangles △APC and △BPC.

1. ∠APC ≅ ∠CPB = 90° ( CP is an altitude),

2. ∠ACP ≅ ∠BCP (CP is a bisector)

3. CP is shared.

Therefore, △APC ≅ △BPC by Angle-Side-Angle.

So AC ≅ BC then the triangle ABC is isosceles.

**Problem 6**

Given:

$\overline{AB} \cong \overline{A_1B_1} $

$\overline{BC} \cong \overline{B_1C_1} $

$\overline{AM} \cong \overline{A_1M_1} $ - medians

Prove that $\triangle ABC \cong \triangle A_1B_1C_1 $

**Solution**

Let's look at the triangles $ABM$ and $A_1B_1M_1$.

1. $\overline{AB} \cong \overline{A_1B_1} $

2. $\overline{AM} \cong \overline{A_1M_1} $

3. $\overline{BC} \cong \overline{B_1C_1}$ and M is a midpoint so $\overline{BM} \cong \overline{B_1M_1}$

$\Rightarrow$ $\triangle ABM \cong \triangle A_1B_1M_1$ by Side-Side-Side.

Therefore, $\angle ABC \cong \angle A_1B_1C_1$

Let's look at $\triangle ABC$ and $\triangle A_1B_1C_1$

1. $AB \cong A_1B_1$

2. $BC \cong B_1C_1$

3. $\angle ABC \cong \angle A_1B_1C_1$

Therefore, the triangles $ABC$ and $A_1B_1C_1$ are congruent by Side-Angle-Side.