ABC is an acute-angled triangle .AP is the diameter of the circumcircle of the triangle ABC, EB and CF are perpendiculars on AC and AB respectiely and they intersect each other at the point other at the point Q. Prove that BPCQ is a parallelogram.

Explanation

`Delta ABC` is an acute angled `BE bot AC and CF bot AD` and Beand CF intersect each, other at a point Q. AP is the diameter of the circumcricle of `Delta ABC`.
To Prove : AP is diameter of the circle `:. angle ABP and angleACP` are bto semi-circular angles.
`:. angle ABP= angle ACP=90^(@).......(1)`
`:. CF bot AB, :. angle CEB=90^(@)`
`:. angle FBQ=90^(@)- angleFQB .........(2)`
Again, `BE bot AC, :. angle BEC=90^(@), :. angleBCQ=90^(@)-angleCQE......(3)`
But `angle FQB= angle CQE[ :.` Opposite angles]
`:.` from (2) and (3) , we get `angle FBQ= angle ECQ`
Then `angle PBQ= angle PBA- angle FBQ= angle ACP- angle ECQ [ :. angle PBA= angle AC and angleFBQ= angle ECQ]= angle PCQ`
`:.` two opposite angles `angle PBQ and angle PCQ` of the quadrilaterl BPCQ are equal.
Hence by the propery of parallelogram. BPCQ is a parallelogram.