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In the adjoining figure, $\triangle ABE \cong \triangle ACD$. Prove that :
(i) $\triangle ADE \sim \triangle ABC$
(ii) $\triangle BOD \sim \triangle COE$
(i) $\triangle ADE \sim \triangle ABC$
(ii) $\triangle BOD \sim \triangle COE$
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(i) As $\triangle ABE \cong \triangle ACD$
$AE = AD, AB = AC$ (CPCT) (1/2 Mark)
Getting $\frac{AD}{AB} = \frac{AE}{AC}$ (1 Mark)
and $\angle A = \angle A$ (common) (1/2 Mark)
$\therefore \triangle ADE \sim \triangle ABC$ (by SAS Similarity) (1/2 Mark)
(ii) As $\triangle ABE \cong \triangle ACD$
$\angle ABE = \angle ACD$ (CPCT) (1 Mark)
and $\angle BOD = \angle COE$ (Vertically opposite angles) (1 Mark)
$\therefore \triangle BOD \sim \triangle COE$ (by AA Similarity) (1/2 Mark)
$AE = AD, AB = AC$ (CPCT) (1/2 Mark)
Getting $\frac{AD}{AB} = \frac{AE}{AC}$ (1 Mark)
and $\angle A = \angle A$ (common) (1/2 Mark)
$\therefore \triangle ADE \sim \triangle ABC$ (by SAS Similarity) (1/2 Mark)
(ii) As $\triangle ABE \cong \triangle ACD$
$\angle ABE = \angle ACD$ (CPCT) (1 Mark)
and $\angle BOD = \angle COE$ (Vertically opposite angles) (1 Mark)
$\therefore \triangle BOD \sim \triangle COE$ (by AA Similarity) (1/2 Mark)