Let $ABC$ be an isosceles triangle in which $AB = AC$. If $D, E, F$ are the mid-points of the sides $BC, CA$ and $AB$ respectively, show that the segment $AD$ and $EF$ bisect each other at right angles.

Given:

$ABC$ is an isosceles triangle in which $AB = AC$. If $D, E, F$ are the mid-points of the sides $BC, CA$ and $AB$ respectively.

To do:

We have to show that the segment $AD$ and $EF$ bisect each other at right angles.

Solution:

$AD$ and $EF$ are joined intersecting at $O$.

Join $DE$ and $DF$.

$D, E$ and $F$ are the mid-points of the sides $BC, CA$ and $AB$ respectively.

This implies,

$AFDE$ is a parallelogram.

Therefore,

$AF = DE$ and $AE = DF$

$AF = AE$           ($E$ and $F$ are mid-points of equal sides $AB$ and $AC$)

This implies,

$AF = DF = DE = AE$

$AFDE$ is a rhombus.

The diagonals of a rhombus bisect each other at right angles.

Therefore,

$AO = OD$ and $EO = OF$

Hence, $AD$ and $EF$ bisect each other at right angles.

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Updated on: 10-Oct-2022

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