Question

If D is any point on the base BC produced, of an isosceles triangle ABC, prove that AD > AB. [3 MARKS]

Answer 1

Process : 2 Marks
Proof : 1 Mark

In $△$ ABC, we have

AB = AC

$\Rightarrow$ $\angle$ABC = $\angle$ACB .....(i)

[ $\because$ Angle opp. to equal sides are equal]

In $△$ ABD, we have

Ext $\angle$ABC > $\angle$ADB

$\Rightarrow$ $\angle$ABC > $\angle$ADB ......(ii)

From (i) and (ii), we get

$\angle$ACB>$\angle$ADB $\Rightarrow$ $\angle$ACD>$\angle$ADC

[ $\because$ $\angle$ACB=$\angle$ACD,$\angle$ADB=$\angle$ADC]

$\Rightarrow$ AD > AC

[$\because$ Side opp. to greater angle is larger]

$\Rightarrow$ AD > AB [$\because$ AB = AC]