In the following figure , the line ABCD is perpendicular to PQ where P and Q are the centres of the circle show that : (ii) AC=BD

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Solution

In the circle with centre $Q, Q O ⊥ A D$
$∴ O A = O D . . . (1)$
(perpendicular drawn from the centre of a circle to a
chord bisect it)
In the circle with centre $P, P O ⊥ B C$
$∴ O B = O C . . . (2)$
(perpendicular drawn from the centre of a circle to a
chord bisect it)
$e q . (1) - (2)$ gives,
$O A - O B = O D - O C$
$A B = C D - - - (3)$
Adding $B C$ to both sides of equation $(3)$
$A B + B C = C D + B C$
$\Rightarrow A C = B D$

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Similar questions

In the following figure, AC and BD are diameters of the circle with centre O. The quadrilateral ABCD is a ___.

AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.

A C

B D

A C

B D

A B C D

In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:

(i) ar (DOC) = ar (AOB)

(ii) ar (DCB) = ar (ACB)

(iii) DA || CB or ABCD is a parallelogram.

[Hint: From D and B, draw perpendiculars to AC.]

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