Question

In $\Delta ABC,AB=AC.D,E$ and $F$ are mid-points of the sides $BC,CA$ and $AB$ respectively. then, $:$ $AD$ is perpendicular to $FC.$

If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

Given: $△ABC$, $AB=AC$, D, E and F are mid points of BC, AB and AC
To prove: $\angle AMF={90}^{\circ }$
Construction: Join ED and FD
In $△ABC$,
$AB=AC$
$\frac{1}{2}AB=\frac{1}{2}AC$
$AE=AF$ ...(I)
Now, E is mid point of AB and F is mid point of AC. By mid point theorem,
$EF\parallel BC$ and $Ef=\frac{1}{2}BC$
Similarly, E is mid point of AB and D is mid point of BC
Thus, $ED=\frac{1}{2}AC$ and $ED\parallel AC$
or $ED=AF$ and $ED\parallel AF$
Similarly, $DF\parallel AE$ and $DF=AE$
Hence, $AFDE$ is a parallelogram.
In a parallelogram opposite sides are equal. Also,
$AE=AF$ ...(I)
Hence, $AFDE$ is a rhombus.
Diagonals of rhombus bisect each other at right angles.
Thus, $AD\perp EF$