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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.
Solution
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.
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