The exterior angles B and C in ΔABC are bisected to meet at a point P. Prove that ∠BPC = 90° − $\dfrac{A}{2}$. Is ABPC a cyclic quadrilateral ?

We know that,

An exterior angle of a triangle equals to the sum of the two interior opposite angles.

∠CBD = ∠A + ∠C

∠BCE = ∠A + ∠B.

Since BP bisects the exterior angle ∠CBD :

∠PBC = $\dfrac{1}{2}$ ∠CBD

Similarly, since CP bisects the exterior angle ∠BCE :

∠PCB = $\dfrac{1}{2}$ ∠BCE

In △BPC,

By angle sum property of triangle,

$\Rightarrow ∠PBC + ∠PCB + ∠BPC = 180^{\circ} \\[1em] \Rightarrow \dfrac{1}{2}∠CBD + \dfrac{1}{2}∠BCE + ∠BPC = 180^{\circ} \\[1em] \Rightarrow \dfrac{1}{2}(∠A + ∠C) + \dfrac{1}{2}(∠A + ∠B) + ∠BPC = 180^{\circ} \\[1em] \Rightarrow ∠A + \dfrac{1}{2}(∠C + ∠B) + ∠BPC = 180^{\circ} \\[1em] \Rightarrow \dfrac{1}{2}(∠B + ∠C) + ∠BPC = 180^{\circ} - A \\[1em] \Rightarrow \dfrac{1}{2}(180^{\circ} - A) + ∠BPC = 180^{\circ} - A \\[1em] \Rightarrow 90^{\circ} - \dfrac{A}{2} + ∠BPC = 180^{\circ} - A \\[1em] \Rightarrow ∠BPC = 180^{\circ} - A - \Big[90^{\circ} - \dfrac{A}{2}\Big] \\[1em] \Rightarrow ∠BPC = 180^{\circ} - A - 90^{\circ} + \dfrac{A}{2}\\[1em] \Rightarrow ∠BPC = 90^{\circ} - \dfrac{A}{2}.$

For A, B, P, C to be concyclic we would need ∠A + ∠BPC = 180° But:

∵ ∠A + ∠BPC = ∠A + $90^{\circ} - \dfrac{∠A}{2} = 90^{\circ} + \dfrac{∠A}{2}$

Hence, proved that $∠BPC = 90° - \dfrac{∠A}{2}$ and ABPC is not cyclic.