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 backward bisects the opposite side.

Answer 1

Let ABCD be a cyclic quadrilateral such that its diagonals AC and BD intersect in P at right angles.
Let PL ⊥ AB such that LP produced to meet CD in M. We have to prove that M bisects CD. i.e.,

Consider arc AD, Clearly, it makes angles ∠ 1 and ∠2 in the same segment.
∠ 1 = ∠ 2                              ...(i)

In the right-angled triangle PLB, we have
∠ 2  + ∠ 3 + ∠ PLB = 180°
⇒  ∠ 2  + ∠ 3 + 90° = 180°
⇒  ∠ 2  + ∠ 3 = 90°          ....(ii)

Since, LPM is a straight line.
∴ ∠ 3 + ∠ BPD + ∠ 4 = 180°
⇒  ∠ 3 + 90° + ∠ 4 = 180°
⇒  ∠ 3 + ∠ 4 = 90°         ...(iii)

From (ii) and (iii), we get
∠ 2  + ∠ 3 = ∠ 3 + ∠ 4 
∠ 2  = ∠ 3                   ...(iv)

From (i) and (iv), we get
∠ 1 = ∠ 4
PM = CM
Similarly,
PM = DM
Hence, CM = MD
Hence proved.