Question

In Fig. AB > AC. Show that AB > AD. [4 MARKS]

Answer 1

Concept : 1 Mark
Process : 2 Marks
Proof : 1 Mark

In $△$ABC, we have

AB $>$ AC [Given]

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

[$\because$ Angle opp. to larger side is greater]

Now, in $△$ACD, CD is produced to B, forming an ext$\angle$ADB.

$\therefore$$\angle$ADB $>$ $\angle$ACD

[$\because$ Exterior angle of $△$ is greater than each of int erior opp.angle]

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

[$\therefore$ $\angle$ACD = $\angle$ACB]

From (i) and (ii), we get

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

$\Rightarrow$ $\angle$ADB $>$ $\angle$ABD [ $\because$ $\angle$ABC = $\angle$ABD]

$\Rightarrow$ AB $>$ AD

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