If two equal chords of a circle intersect each other, then prove that ordered parts of one chord are respectively equal to the corresponding parts of other chord.

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Solution

Given :
Let on a circle $C (O, r)$ two equal chords $A B$ and $C D$ intersect each other at point $P$ .
To Prove :
$A P = C P$
and $P B = P D$
Construction : Draw $O L ⊥ A B$ and $O M ⊥ C D$ and join $O P$
Proof : $O L ⊥ A B$ and $O M ⊥ C D$
$O L = O M$ …..(i) $[∵ A B = C D]$ (Equal chords are (RBSESolutions.com) equidistant from center)
In right angled $Δ O L P$ and $Δ O M P$
$O L = O M$ [from eqn (i)]
$∠ O L P = ∠ O M P$ [each $90^{0}$ ]
$O P = O P$ [common side]
$∴ Δ O L P = Δ O M P$
$L P = P M$ …..(ii)
But $A B = C D$
$\frac{1}{2} A B = \frac{1}{2} C D$
$A L = C M$ ……(iii)
Adding equations (ii) and (iii)
$L P + A L = P M + C M$
$A P = P C$ ……(iv)
thus $A B = C D . . . . (v)$
$A P = C P)$
Now subtracting equation (iv) from (v)
$A B - A P = C D - C P$
$P B = P D$

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