Question

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

Answer 1

AB Is greater than AC (given)
LABC is greater than LACB

As LB AND LC are bisected ....
LDBP= LPBC and LPCB= LECP

As LACB is greater than LABC
LPBC is greater than LPCB
therefore PC is greater than PB
HENCE PROVED....


Mark as brainliest

Answer 2

[\tex]\huge\underline\mathbb\red{ANSWER-}

Let,

∠ABC=x and ∠ACB=y

If AB>AC

then,

∠y >∠x (angle side theorem)

So the external angle will be,

∠180 − x , ∠180 - y for x and y respectively.

So the angle made by Pbwith B and C will be

∠PBC = ∠( 2 / 180−x)

∠PCB =∠( 2 / 180−y)

Since ∠y > ∠x,

∠180 − y < ∠180 - x

So,

∠PCB < ∠PBC

So,

PB < PC...