## In the figure, D,E and F are the mid points of sides AB, AC and BC respectively and AG is perpendicular on BC. Prove that DEFG is a cyclic quadrilateral.

Apr 15, 2018

Given:

In $\Delta A B C$

$D , E , F$ are midpoints of $A B , A C \mathmr{and} B C$ respectively and $A G \bot B C$.

Rtp:

Proof:

As $D , E , F$ are midpoints of $A B , A C \mathmr{and} B C$ respectively,
By midpoints theorem of a triangle we have

$D E \text{||} B C \mathmr{and} G F \mathmr{and} D E = \frac{1}{2} B C$

Similarly

$E F \text{||} A B \mathmr{and} E F = \frac{1}{2} A B$

Now in $\Delta A G B , \angle A G B = {90}^{\circ}$ Since $A G \bot B C$ given.

So $\angle A G B = {90}^{\circ}$ will be semicircular angle of the circle drawn taking AB as diameter i,e centering D,

Hence $A D = B D = D G \implies D G = \frac{1}{2} A B$

So in quadrilateral $D E F G$

$D G = E F \mathmr{and} D E \text{||"GF}$

This means the quadrilateral $D E F G$ is an isosceles trapezium which must be cyclic one,