In Delta ABC,P,Q and R are the midpoints of A | vedclass

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(B) In $\Delta ABC$,$P$ is the midpoint of $AB$ and $R$ is the midpoint of $AC$. By the Midpoint Theorem,$PR \parallel BC$ and $PR = \frac{1}{2} BC =

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$PQR$

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(B) In $\Delta ABC$,$P$ is the midpoint of $AB$ and $R$ is the midpoint of $AC$. By the Midpoint Theorem,$PR \parallel BC$ and $PR = \frac{1}{2} BC = BQ = QC$.Similarly,$Q$ is the midpoint of $BC$ and $R$ is the midpoint of $AC$,so $QR \parallel AB$ and $QR = \frac{1}{2} AB = AP = PB$.Also,$P$ is the midpoint of $AB$ and $Q$ is the midpoint of $BC$,so $PQ \parallel AC$ and $PQ = \frac{1}{2} AC = AR = RC$.Thus,in quadrilateral $APQR$,$AP \parallel QR$ and $AR \parallel PQ$,making $APQR$ a parallelogram.In a parallelogram,opposite angles are equal. Therefore,$\angle BAC = \angle PQR$.

In $\Delta ABC$,$P$,$Q$ and $R$ are the midpoints of $AB$,$BC$ and $CA$ respectively,then $\angle BAC = \angle \ldots \ldots \ldots$

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