Question:
The diagonals of a cyclic quadrilateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.
Solution:
Let ABCD be a cyclic quadrilateral whose diagonals AC and BD intersect at O at right angles.
Let OL ⊥ AB such that LO produced meets CD at M.
Then we have to prove that CM = MD
Clearly, ∠1 = ∠2 [Angles in the same segment]
∠2 + ∠3 = 90° [∵ ∠OLB = 90°]
∠3 + ∠4= 90° [∵ LOM is a straight line and ∠BOC = 90°]
∴ ∠2 + ∠3 = ∠3 + ∠4 ⇒∠2 = ∠4
Thus, ∠1 = ∠2 and ∠2 = ∠4 ⇒ ∠1 = ∠4
∴ OM = CM and, similarly, OM = MD
Hence, CM = MD