Question
In isosceles $\triangle ABC, AB = AC.$ The side $BA$ is produced to $D$ such that $BA = AD.$Prove that$: \angle BCD = 90^\circ $
✓
Answer
Const: Join $CD.$
In $\triangle ABC,$
$AB = AC .........[$ Given $]$
$\therefore \angle C = \angle B .......(i) [$angles opp. to equal sides are equal$]$
In$ \triangle ACD,$
$AC= AD ...[$Given$]$
$\therefore \angle ADC = \angle ACD ........(ii)$
Adding $(i)$ and $(ii)$
$\angle B + \angle ADC = \angle C + ACD$
$\angle B + \angle ADC = \angle BCD ....(iii)$
In $\triangle BCD,$
$\angle B + \angle ADC + \angle BCD = 180^\circ $
$\angle BCD + \angle BCD = 180^\circ .......[$From $(iii)]$
$2\angle BCD = 180^\circ $
$\angle BCD = 90^\circ $
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