The diagonals of a cyclic quadrilateral are at right triangles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.

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Consider ABCD as a cyclic quadrilateral with diagonal AC and BD intersecting at right anglesWe know that OL⊥AB so that ∠O meets the line CD at the point MFrom the figure, we know that the angles in the same segment are equal∠1=∠2We know that ∠OLB=90o we get ∠2+∠3=90o.....(1)We know that OLM is a straight line and ∠BOC=90o So we get ∠3+∠4=90o.....(2)It can be written as ∠2+∠3=∠3+∠4On further calculation ∠2=∠4So we get ∠1=∠2 and ∠2=∠4It can be written as ∠1=∠4We get OM=CMIn the same wayOM=MDso CM=MDtherefore, it is proved that the perpendicular from the point of their intersection on any side when produced backward, bisects the opposite side.

Question Text	The diagonals of a cyclic quadrilateral are at right triangles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.
Answer Type	Text solution:1
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The diagonals of a cyclic quadrilateral are at right triangles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.

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