`Delta`ABC is an acute angle triangle inscribed in a circle in a circle .AD is a diameter of the circle . Two perpendiculars BE and CF are drawn from B and C to AC and AB respectively , which intersect each other at the point G . Prove that BDCG is a parallelogram.

Explanation

Let `DeltaABC` is an acute angled traigle triangle , insribed in a circle with center at O .Two perpendiculars BE and CF are drawn from B and C to the sides AC and AB which intersect each other at the point G. AD is a diameter of the circle .
To Prove : BDCG is parallelogram .
Proof : AD is a diameter , `thereforeangleABD=90^(@)" " [because` semicircular angle]
Again two angles in circle produced by the chord BD are `angleBCDandandBAD` .
`thereforeangleBCD=angleBAD" " [because` angles in the same segment of circle are equal.]
`=90^(@)-angleBDA[becauseangleABD=90^(@),becauseangleBAD+angleBDA=90^(@)]`
=`90^(@)-angleACB[becauseangleBDAand angleACB` are angles in circle produced by the arc AB.]
`=angleCBE[becauseangleBEC=90^(@),angleCBE+angleBCE=90^(@)] =angleCBG"i.e", angleBCD=angleCBG`
i.e. , `angleBCD=angleCBG`.
But these are alternate angles , `therefore` BG||DC
Similarly , it can be proved that BD||GC .So , two opposite side of the quadrilateral BDCG are parallel.
Hence BDCG is a paralleogram