Answer:
See Below.
Step-by-step 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:
[tex]\angle PAB+\angle PBA+\angle PCB+\angle PBC=180[/tex]
Since they are all equivalent:
[tex]4x=180[/tex]
Hence:
[tex]x=45^\circ[/tex]
∠ABC is the sum of ∠PBA and ∠PBC, each of which measures 45°. Hence:
[tex]\angle ABC=\angle PBA+\angle PBC=45+45=90^\circ[/tex]
