Question

Sides AB and AC of ∆ ABC are extended to points P and Q respectively. Show that 
AC > AB, if ∠PBC <∠QCB.

Answer 1

Angle PBC being the exterior angle of triangle ABC.

By the property that the exterior angle is the sum of two interior opposite angles, we can say that,

PBC = A + ACB

By the same property we can say that,

QCB = A + ABC

The above two equations we can say that,

PBC < QCB

Now, we know that side opposite to greater angle is longer.

Therefore, ACB< ABC

Also, AC > AB