Question

Prove ΔPAB is isosceles.

Answer 1

Answer:

See explanation

Step-by-step explanation:

If $PX\cong PY,$ then triangle PXY is isosceles triangle. Angles adjacent to the base XY of an isosceles triangle PXY are congruent, so

$\angle 1\cong \angle 2$

and

$m\angle 1=m\angle 2$

Angles 1 and 3 are supplementary, so

$m\angle 3=180^{\circ}-m\angle 1$

Angles 2 and 4 are supplementary, so

$m\angle 4=180^{\circ}-m\angle 2$

By substitution property,

$m\angle 4=180^{\circ}-m\angle 2=180^{\circ}-m\angle 1=m\angle 3$

Hence,

$\angle 3\cong \angle 4$

Consider triangles APX and BPY. In these triangles:

• $PX\cong PY$ - given;
• $\angle 5\cong \angle 6$ - given;
• $\angle 3\cong \angle 4$ - proven,

so $\triangle APX\cong \triangle BPY$ by ASA postulate.

Congruent triangles have congruent corresponding sides, then

$AP\cong BP$

Therefore, triangle APB is isosceles triangle (by definition).