Question

given that ∠w≅∠x and that ∠y≅∠z, prove each of the following: a. △abe≅△ace b. (overline{ab}congoverline{ac}) c. (overline{ad}perpoverline{bc}) for ease of scoring, please indicate the beginning of each proof by typing a., b., or c.

Explanation:

A.

Step1: Identify common side

In $\triangle ABE$ and $\triangle ACE$, $AE = AE$ (common side).

Step2: Use given angle - congruences

We know that $\angle w\cong\angle x$ and $\angle y\cong\angle z$. By the Angle - Side - Angle (ASA) congruence criterion, since $\angle w=\angle x$, $AE = AE$, and $\angle y=\angle z$, we have $\triangle ABE\cong\triangle ACE$.

B.

Step1: Use triangle congruence result

Since $\triangle ABE\cong\triangle ACE$ (proven in part A), corresponding parts of congruent triangles are congruent. So, $\overline{AB}\cong\overline{AC}$ (corresponding sides of congruent triangles $\triangle ABE$ and $\triangle ACE$).

C.

Step1: Use side - congruence from part B

We have $\overline{AB}\cong\overline{AC}$. Also, $\triangle ABE\cong\triangle ACE$ implies $\angle BAD=\angle CAD$.

Step2: Consider $\triangle ABD$ and $\triangle ACD$

In $\triangle ABD$ and $\triangle ACD$, $\overline{AB}\cong\overline{AC}$, $\angle BAD = \angle CAD$, and $AD=AD$ (common side). So, $\triangle ABD\cong\triangle ACD$ by the Side - Angle - Side (SAS) congruence criterion.

Step3: Use angle - congruence from $\triangle ABD$ and $\triangle ACD$

Since $\triangle ABD\cong\triangle ACD$, $\angle ADB=\angle ADC$. And since $\angle ADB+\angle ADC = 180^{\circ}$ (linear - pair of angles), $\angle ADB=\angle ADC = 90^{\circ}$. So, $\overline{AD}\perp\overline{BC}$.

Answer:

A. $\triangle ABE\cong\triangle ACE$ is proven by ASA criterion.
B. $\overline{AB}\cong\overline{AC}$ as corresponding parts of congruent $\triangle ABE$ and $\triangle ACE$.
C. $\overline{AD}\perp\overline{BC}$ is proven by showing $\triangle ABD\cong\triangle ACD$ and using linear - pair of angles.