Question

In the given figoure, D is a point on side BC of ∆ABC such that AD=AC .show that AB>AD.

Answer 1

given=AD=AC
in triangle ABD
                      AB+BD>AD............1
IN TRIANGLE ACD
                             AC+CD>AD............2
FROM 1AND2
AB+BD+AC+CD>AD+AD
AB+BC+AC>AD+AC
AB+BC>AD=AB>AD

Answer 2

Answer:In ∆ABC D is a point on BC such that

AD=AC

Now in ∆ADC, AD=AC

Hence ∠ADC = ∠ACD  (angle opp. to equal sides are equal)

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)