Question

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

Answer 1

Step-by-step explanation:

∠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

thus AC.> AB

Hence proved.

hope it helps