Question

The sides $AB$ and $AC$ of a triangle $ABC$ are produced; and the bisectors of the external angles at $B$ and $C$ meet at $P$. Prove that if $AB>AC$, then $PC>PB$.

Answer 1

Here AB$>$AC

$<ACB><ABC$

Angle opposite to greater side is greater

$<ECB<<DBC$ (linear angle)

$\frac{1}{2}<ECB<\frac{1}{2}<DBC$

$<PCB<<PBC$

$\because PB<PC$

$PC>PB$ Hence proved.