In a $\triangle ABC, E$ and $F$ are the mid-points of $AC$ and $AB$ respectively. The altitude $AP$ to $BC$ intersects $FE$ at $Q$. Prove that $AQ = QP$.

Given:

In a $\triangle ABC, E$ and $F$ are the mid-points of $AC$ and $AB$ respectively. The altitude $AP$ to $BC$ intersects $FE$ at $Q$. 

To do:

We have to prove that $AQ = QP$.

Solution:

Join $EF$.

$AP \perp BC$ is drawn which intersects $EF$ at $Q$ and meets $BC$ at $P$.

In $\triangle ABC$,

$EF \parallel BC$ and $EF = \frac{1}{2}BC$

$\angle F = \angle B$

In $\triangle ABP$,

$F$ is the mid point of $AB$ and $Q$ is the mid point of $FE$

This implies,

$FQ \parallel BC$

$Q$ is the mid point of $AP$.

$AQ = QP$

Hence proved.

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Updated on: 10-Oct-2022

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