Question

In adjacent figure, sides AB of AC of ∆ABC are extended to points P and Q respectively
Also,PBC <QCB. show that AC>AB

Answer 1

By exterior angle property
angle PBC= angle A+ angleACB (eqn.1)
and angle QCB= angle A +angleABC. (eqn.2)
By eqn.1and 2
angleA+ACB<angleA+ABC
It means that
angle ACB< angleABC
then AB<AC. (side opposite to greater angle is greater)

Answer 2

Answer:

Step-by-step explanation:

Angle a plus angle ACB is smaller than angle a plus angle ABC

Angle ACB is smaller than angle ABC

So,AB is smaller than AC

HENCE PROVED