Question

In triangle ABC, AB > AC and D is a point
in side BC. Show that : AB > AD.​

Answer 1

solution;

In ∆ABC the angle opposite to side AB is ∠ACB

And the angle opposite to side AC is ∠ABC

Here AB>AC

Therefore ∠ACB >∠ABC····(1) (since greater side has greater angle opposite to it)

In ∆ADC

∠ADB =∠ACD+∠ DAC

Or ∠ADB= ∠ACB +∠DAC

therefore ∠ADB> ∠ACB······(2)

from(1) and(2) we have∠ADB> ∠ABC ie ∠ADB >∠ABD

Now in ∆ABD ∠ADB >∠ABD

Hence AB >AD (since greater angle has greater side opposite to it)