Answer:
See Below.
Explanation:
In the given figure, AP = BP = PC.
And we want to prove that ∠ABC is a right angle.
Since AP = BP and BP = PC, we can create two isosceles triangles: ΔAPB and ΔCPB.
By the definition of isosceles triangles, in ΔAPB, ∠PAB and ∠PBA are equivalent. Let the measure of each of them be x°.
Likewise, in ΔCPB, ∠PCB and ∠PBC are equivalent.
And since AP = BP = PC, each of the angles∠PCB and ∠PBC will also be equivalent to x°.
And since the sum of the interior angles of a triangle total 180°, we acquire:
![\angle PAB+\angle PBA+\angle PCB+\angle PBC=180](https://img.qammunity.org/2022/formulas/mathematics/high-school/5vzgwk7xdbt36ajgm9y94tsp2xtp9r6tr6.png)
Since they are all equivalent:
![4x=180](https://img.qammunity.org/2022/formulas/mathematics/high-school/ppxwnf5yq20i4cuyynufndz3itsk8dss41.png)
Hence:
![x=45^\circ](https://img.qammunity.org/2022/formulas/mathematics/high-school/8p3v7u4e6gxdbpx9eetb3r1097p1ycpsxm.png)
∠ABC is the sum of ∠PBA and ∠PBC, each of which measures 45°. Hence:
![\angle ABC=\angle PBA+\angle PBC=45+45=90^\circ](https://img.qammunity.org/2022/formulas/mathematics/high-school/c7dz2lnl4fkz0uz6q50cpkddvzdd3ofnj3.png)