Question

Please prove?

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.

Answer 1

Given:

In `Delta ABC`

`D,E,F` are midpoints of `AB,ACand BC` respectively and `AG_|_BC`.

Rtp:

DEFG is a cyclic quadrilateral.

Proof:

As `D,E,F` are midpoints of `AB,ACand BC` respectively,
By midpoints theorem of a triangle we have

`DE"||"BC orGF and DE=1/2BC`

Similarly

`EF"||"AB and EF=1/2AB`

Now in `Delta AGB,angle AGB=90^@` Since `AG_|_BC` given.

So `angle AGB=90^@` will be semicircular angle of the circle drawn taking AB as diameter i,e centering D,

Hence `AD=BD=DG=>DG=1/2AB`

So in quadrilateral `DEFG`

`DG=EF and DE"||"GF"`

This means the quadrilateral `DEFG` is an isosceles trapezium which must be cyclic one,