Question

The diagonals of a cyclic quadrilateral are at right angles. Prove that perpendiculars from thepoint of their intersection on any side when produced backward bisect 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 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 = ∠4 …….(iv)

From (i) and (iv), we get

∠1 = ∠4 …….(v)

⇒ PM = CM

Similarly, PM = DM

Hence, CM = MD

Hence proved.

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