Let P be an interior point of a triangle ABC. | vedclass

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(C) Let $\angle PAB = 	heta$ and $\angle PAC = \phi$. Since $Q$ is the reflection of $P$ in $AB$,we have $AQ = AP$ and $\angle QAB = \angle PAB = 	h

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$90$

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(C) Let $\angle PAB = 	heta$ and $\angle PAC = \phi$. Since $Q$ is the reflection of $P$ in $AB$,we have $AQ = AP$ and $\angle QAB = \angle PAB = 	heta$. Since $R$ is the reflection of $P$ in $AC$,we have $AR = AP$ and $\angle RAC = \angle PAC = \phi$. Given that $Q, A, R$ are collinear,the angle $\angle QAR = 180^{\circ}$. From the figure,$\angle QAR = \angle QAB + \angle BAC + \angle RAC = 	heta + (	heta + \phi) + \phi = 2(	heta + \phi) = 180^{\circ}$. Thus,$	heta + \phi = 90^{\circ}$. Since $\angle BAC = 	heta + \phi$,we have $\angle BAC = 90^{\circ}$.

Let $P$ be an interior point of a $\triangle ABC$. Let $Q$ and $R$ be the reflections of $P$ in $AB$ and $AC$,respectively. If $Q, A, R$ are collinear,then $\angle A$ equals (in $^{\circ}$)

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