Two equal chord $A B$ and $C D$ of a circle with centre $O$ , intersect each other at point $P$ inside the circle. Prove that:
$A P = C P$
$B P = D P$

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Solution

$A B = C D$ ........(A)
$∴ O R = O Q$ ...(3) (equal chords have equal distance from the centre of the circle)
$O P^{2} = O R^{2} + R P^{2}$ ......(1)
$O P^{2} = O Q^{2} + Q P^{2}$ ....(2)
From (1) , (2) and (3)
$O P^{2} = R P^{2}$ ....(4)
$\Rightarrow Q P + R P . . . . . (4)$
$A B = C D$
$\frac{A B}{2} = \frac{C D}{2}$
$Q B = D R$ (perpendicular from centre bisech the chord)
(ii) On adding (4) and (5)
$Q B + Q P = R P + D R \Rightarrow B P = D P$ ......(6)
$(A) - (6)$
(i) $\Rightarrow A B - B P = C D - D P$
$\Rightarrow A P = C P$ ....(7)
$∴ A P = C P$ & $B D = D P$

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