Question

D is the midpoint of any side BC of triangle ABC such that AD=AC show that AB > AD

Answer 1

In ∆ABC D is a point on BC such that

AD=AC

Now in ∆ADC, AD=AC

Hence ∠ADC = ∠ACD

Again ∠ADB(ext. angle)=∠DAC+∠ACD(int. opp. angles)

Therefore ∠ADB > ∠ACD…(1)

In ∆ABD ∠ADC(ext.angle)= ∠ABD+∠BAD(int. opp angles)

⇒∠ADC >∠ABD

Or, ∠ACD>∠ABD….(2) (since ∠ADC=∠ACD)

From (1) and (2) We have ∠ADB>∠ABD

Now in ∆ABD, ∠ADB>∠ABD

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