Question 1170429: Given: \overline{CA}
CA
bisects \angle BAD∠BAD and \angle B \cong \angle D.∠B≅∠D.
Prove: \triangle ABC \cong \triangle ADC△ABC≅△ADC.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's the step-by-step proof:
**Given:**
1. $\overline{CA}$ bisects $\angle BAD$.
2. $\angle B \cong \angle D$.
**Prove:**
$\triangle ABC \cong \triangle ADC$.
**Proof:**
| **Statement** | **Reason** |
|---|---|
| 1. $\overline{CA}$ bisects $\angle BAD$. | Given |
| 2. $\angle BAC \cong \angle DAC$. | Definition of angle bisector. |
| 3. $\angle B \cong \angle D$. | Given |
| 4. $\overline{CA} \cong \overline{CA}$. | Reflexive Property of Congruence. |
| 5. $\triangle ABC \cong \triangle ADC$. | Angle-Side-Angle (ASA) Congruence Theorem. |
**Explanation:**
1. We are given that $\overline{CA}$ bisects $\angle BAD$. This means that $\overline{CA}$ divides $\angle BAD$ into two congruent angles.
2. By the definition of an angle bisector, $\angle BAC$ and $\angle DAC$ are congruent.
3. We are given that $\angle B \cong \angle D$.
4. $\overline{CA}$ is a shared side between $\triangle ABC$ and $\triangle ADC$. By the Reflexive Property of Congruence, any segment is congruent to itself.
5. We have shown that two angles ($\angle BAC$ and $\angle B$) and the included side ($\overline{CA}$) of $\triangle ABC$ are congruent to the corresponding parts ($\angle DAC$, $\angle D$, and $\overline{CA}$) of $\triangle ADC$. Therefore, by the Angle-Side-Angle (ASA) Congruence Theorem, $\triangle ABC \cong \triangle ADC$.
|
|
|
