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Prove ΔPAB is isosceles.

Prove ΔPAB is isosceles.-example-1
User Euralis
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Answer:

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).

User Vinze
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