Question

sides AB and AC of A ABC are
extended to points P and Q respectively. Also,anglePBC<angle QCB. Show that AC>AB.​

Answer 1

Answer:

Answer

∠PBC is the exterior angle of △ABC.

Since exterior angle is sum of interior opposite angles.

So,

∠PBC=∠A+∠ACB

similarly,

∠QCB is the exterior angle of △ABC

So,

∠QCB=∠A+∠ABC

Now,

⇒ ∠PBC<∠QCB

⇒ ∠A+∠ABC<∠A+∠ACB

⇒ ∠ABC<∠ACB

On writing the opposite sides of the angles, we get

⇒ AB<AC

Hence proved.