In the figure, $BE \perp AC$. $AD$ is any line from $A$ to $BC$ intersecting $BE$ in $H. P, Q$ and $R$ are respectively the mid-points of $AH, AB$ and $BC$. Prove that $PQR = 90^o$.
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Given:

In $\triangle ABC, BE \perp AC$.

$AD$ is any line from $A$ to $BC$ meeting $BC$ in $D$ and intersecting $BE$ in $H. P, Q$ and $R$ are respectively mid points of $AH, AB$ and $BC$.

To do:

We have to prove that $PQR = 90^o$.

Solution:

Join $PQ$ and $QR$.

In $\triangle ABC$,

$Q$ and $R$ the mid points of $AB$ and $BC$.

This implies,

$QR \parallel AC$ and $QR = \frac{1}{2}AC$

Similarly,

In $\triangle ABH$,

$Q$ and $P$ are the mid points of $AB$ and $AH$

This implies,

$QP \parallel BE$

$AC \perp BE$

Therefore,

$QP \perp QR$

This implies,

$\angle PQR = 90^o$.

Hence proved.

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Updated on: 10-Oct-2022

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