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

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

If 2 angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.

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

If 2 angles and the nonincluded side of one triangle are congruent to two angles and the nonincluded side of another triangle, then the triangles are congruent.

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

If the sides two triangles are congruent, then both triangles are congruent.

## Problems

Problem 1
Given ABC - an isosceles triangle.
AM and BN are angle bisectors.
Prove that AMBN.

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 AMBN.

Problem 2
Given:
ABC is a triangle.
CM is a median.
AA1 ⊥ CM and BB1 ⊥ CM.
Prove that AA1BB1

Solution:
Triangles AA1M and BB1M are congruent (by Angle-Side-Angle) because
∠BB1M ≅ ∠AA1M = 90°,
∠AMA1 ≅ ∠BMA1,
AMBM
Therefore, AA1BB1.

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 OPOQ 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 ACBC 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 ACBC 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.

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