Question

ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at right angles at M. Prove that any line passing through M and bisects any side of the quadrilateral is perpendicular to the opposite side.

Answer 1

Answer:

Step-by-step explanation:

Consider ABCD be a cyclic quadrilateral such that its diagonals AC and BD intersect in O at right angles. Let OL ⊥ AB such that LO produced meet CD in M.

To prove that M bisects CD i.e., CM = MD.

Consider are AD. Clearly, it makes angles ∠x and ∠y in the same segment.

∠x = ∠y …........... (1)

In right triangle OLB, we have

∠y + ∠r + ∠OLB = 180°

⇒ ∠y + ∠r + 90° = 180°

⇒∠y + ∠r = 90° ... (2)

Since LPM is a straight line.

∠r + ∠BOC + ∠s = 180°

⇒ ∠r + 90° + ∠s= 180°

⇒∠r + ∠s = 90° ... (3)

From (2) and (3) we get

∠y + ∠r = ∠r + ∠s ⇒ ∠y = ∠s

From (1) and (4) we get

∠x = ∠s

⇒ OM = DM

Similarly, OM = DM

Hence, CM = MD

Answer 2

Answer:

the answer can be proved directly by bhramagubta theorem