Question

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

Answer 1

In $\Delta ABC,$ $AB=AC,D,E$ and $F$ are the midpoints of the sides $BC,CA$ and $AB$ respectively,

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

To prove : $AD$ and $EF$ bisect each other at right angles.

Construction: Join $DE$ and $DF.$

Proof : $D,E$ and $F$ are the midpoints of the sides $BC,CA$ and $AB$ respectively.

$FD\parallel AC,FD=\frac{1}{2}AC$ [By Midpoint theorem]

Similarly, $DE\parallel AB,DE=\frac{1}{2}AB$

$\therefore$ $AFDE$ is a $\parallel$gram

$\therefore$ $AF=DE$ and $AE=DF$

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

$\therefore$ $AF=DF=DE=AE$

$\therefore$ $AFDE$ is a rhombus

$\because$ The diagonals of a rhombus bisect each other at right angle.

$\therefore$ $AO=OD$, $EO=OF$ and, $AD\perp FE$

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