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State AA criterion of similarity of two triangles and use it to prove the following.
In the given figures of $\triangle ABC$ and $\triangle PQR$, $AD$ and $PS$ are angle bisectors of $\angle BAC$ and $\angle RPQ$ respectively. If $\triangle ABC \sim \triangle PQR$, prove that $\triangle ACD \sim \triangle PRS$.
In the given figures of $\triangle ABC$ and $\triangle PQR$, $AD$ and $PS$ are angle bisectors of $\angle BAC$ and $\angle RPQ$ respectively. If $\triangle ABC \sim \triangle PQR$, prove that $\triangle ACD \sim \triangle PRS$.
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Solution: Statement: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
Proof: $\triangle ABC \sim \triangle PQR$ (given)
As corresponding angles of similar triangles are equal.
$\therefore \angle A = \angle P, \angle B = \angle Q, \angle C = \angle R$ -------(i)
In $\triangle ACD$ and $\triangle PRS$
$\frac{1}{2} \angle BAC = \frac{1}{2} \angle RPQ$ (As $AD$ and $PS$ are angle bisectors)
$\angle CAD = \angle RPS$
$\angle C = \angle R$ (from (i))
$\therefore \triangle ACD \sim \triangle PRS$ (by AA similarity criterion)
Proof: $\triangle ABC \sim \triangle PQR$ (given)
As corresponding angles of similar triangles are equal.
$\therefore \angle A = \angle P, \angle B = \angle Q, \angle C = \angle R$ -------(i)
In $\triangle ACD$ and $\triangle PRS$
$\frac{1}{2} \angle BAC = \frac{1}{2} \angle RPQ$ (As $AD$ and $PS$ are angle bisectors)
$\angle CAD = \angle RPS$
$\angle C = \angle R$ (from (i))
$\therefore \triangle ACD \sim \triangle PRS$ (by AA similarity criterion)