Question

$\Delta$ ABC is an isosceles triangle with AB = AC. Side BA is produced to D such that AB = AD. Prove that $\angle$BCD is right angle. [4 MARKS]

Answer 1

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

Given: A $\Delta$ ABC such that AB = AC. Side BA is produced to D such that AB = AD.

Construction: Join CD.

To Prove: $\angle$BCD = ${90}^{\circ }$

Proof: In $\Delta$ ABC, we have AB = AC
$\Rightarrow$ $\angle$ACB = $\angle$ABC.....(i)
Now, AB = AD [Given]
$\therefore$, AD = AC [$\because$ AB=AC]


Thus, in $\Delta$ADC, we have
AD = AC
$\Rightarrow$ $\angle$ACD = $\angle$ADC......(ii) [Angles opp. to equal sides are equal]

Adding (i) and (ii), we get
$\angle$ACB+$\angle$ACD = $\angle$ABC+$\angle$ADC
$\Rightarrow$ $\angle$BCD = $\angle$ABC+$\angle$BDC [$\because$ $\angle$ADC = $\angle$BDC]

$\Rightarrow$ $\angle$BCD+$\angle$BCD = $\angle$ABC+$\angle$BCD [Adding $\angle$BCD on both sides]


$\Rightarrow$ 2$\angle$BCD = ${180}^{\circ }$ [Sum of the angles of a $\Delta$ is ${180}^{\circ }$]
Hence,$\angle$BCD is a right angle