Question

In the given figure $△ABC$ is isosceles with $AB=AC$.$D,E$ and $F$ are respectively the mid-points of $BC,CA$ and $AB$.Show that the line segment $AD$ is perpendicular to the line segment $EF$ and is bisected by it.

Answer 1

R.E.F image

Given : $△$ ABC is isosceles with AB=AC ,E and F are the mid-points of BC, CA and AB

To prove: AD$\perp EF$and is bisected by t

construction: Join D, F and F

Proof: $DE\left|\right|AC$ and $DE=\frac{1}{2}AB$

and $DF\left|\right|Ac$ and$DE=\frac{1}{2}AC$

The line segment joining midpoints of two sides of a triangle is parallel to the third side and is half of it

DE = DF $(\because AB=AC)$ Also AF=AE

$\therefore AF=\frac{1}{2}AB,AE=\frac{1}{2}AC$

$\therefore DE=AE=AF=DF$

and also DF$\left|\right|$ AE and $DE\left|\right|AF$

$\Rightarrow$ DEAF is a rhombus.

since diagrams of a rhombus bisect each other of right angles

$\therefore AD\perp EF$ and is bisected by it