Show that P is equidistant from B and C.

Question

P is drawn using the exterior angle bisector of A.

Answer 1

let $\theta=\frac \alpha 2$ and extend $PA$

$$\angle PAB = \theta \text{ ( Opposite angles at A)}$$ $$\angle PCB =\angle PAB = \theta \text{ ( both subtend PB )}$$

Let $x=\angle APC$ and $y=\angle ACP$

Then

$$x+y=\theta \text{ ( exterior angle in }\triangle APC \;) $$

$$\angle APC=\angle ABC = x \text{ ( both subtend AC )}$$

$$\angle ACP=\angle ABP = y \text{ ( both subtend AP )}$$

$$\angle PBC =\angle PBA+\angle ABC = x+y=\theta$$

So $\triangle PBC$ is isosceles ( $\angle PBC= \angle PCB=\theta$ )