Question
In the given figure, $\text{AB}=\text{AC}$ and $\angle DBC=\angle ECB=90^{\circ}$
Prove that:
(i) $BD = CE$
(ii) $AD = AE$
Prove that:
(i) $BD = CE$
(ii) $AD = AE$
✓
Answer
In $\triangle ABC,$
$\text{AB} = \text{AC} ......[$Given$]$
$\therefore \angle ACB = \angle ABC ......[$angles opp. to equal sides are equal$]$
$\Rightarrow \angle ABC = \angle ACB .....(i)$
$\angle DBC =\angle ECB = 90^\circ ......[$Given$]$
$\Rightarrow \angle DBC = \angle ECB ….(ii)$
Subtracting $(i)$ from $(ii)$
$\angle DCB − \angle ABC = \angle ECB − \angle ACB$
$\Rightarrow \angle DBA = \angle ECA ........(iii)$
In $\triangle DBA$ and $\triangle ECA,$
$\angle DBA = \angle ECA ......[$From $(iii)]$
$\angle DAB = \angle EAC .......[$Vertically opposite angles$]$
$\text{AB} = \text{AC} ......[$Given$]$
$\therefore \triangle DBA ≅ \triangle ECA .......[\text{ASA}]$
$\Rightarrow \text{BD} = \text{CE} ...[$c. p. c. t$]$
Also,
$\text{AD} = \text{AE} ...[$c. p. c. t$]$
$\text{AB} = \text{AC} ......[$Given$]$
$\therefore \angle ACB = \angle ABC ......[$angles opp. to equal sides are equal$]$
$\Rightarrow \angle ABC = \angle ACB .....(i)$
$\angle DBC =\angle ECB = 90^\circ ......[$Given$]$
$\Rightarrow \angle DBC = \angle ECB ….(ii)$
Subtracting $(i)$ from $(ii)$
$\angle DCB − \angle ABC = \angle ECB − \angle ACB$
$\Rightarrow \angle DBA = \angle ECA ........(iii)$
In $\triangle DBA$ and $\triangle ECA,$
$\angle DBA = \angle ECA ......[$From $(iii)]$
$\angle DAB = \angle EAC .......[$Vertically opposite angles$]$
$\text{AB} = \text{AC} ......[$Given$]$
$\therefore \triangle DBA ≅ \triangle ECA .......[\text{ASA}]$
$\Rightarrow \text{BD} = \text{CE} ...[$c. p. c. t$]$
Also,
$\text{AD} = \text{AE} ...[$c. p. c. t$]$
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