Question

Prove ΔPAB is isosceles.

Answer 1

See explanation

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).