If two equal chords of a circle intersect, then prove that their segments are equal. [3 MARKS]

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Solution

Concept: 1 Mark
Application: 2 Marks

Given: A circle with center O. Its two equal chords AB and CD intersect at E.

To prove: AE = DE and CE =BE

Construction: Draw $O M ⊥ A B$ and $O N ⊥ C D$ . Join OE.

Proof: In $Δ O M E$ and $Δ O N E$

$O M = O N$ (Equal chords of a circle are equidistant from the centre)

$O E = O E$ (Common)

$∴$ $Δ O M E ≅ Δ O N E$ (R.H.S)

$\Rightarrow M E = N E$ (C.P.C.T)

$\Rightarrow A M + M E = D N + N E (∵ A M = D N = \frac{1}{2} A B = \frac{1}{2} C D)$
[Perpendicular from centre divide the chord into two equal parts]

$\Rightarrow A E = D E$

$\Rightarrow A B - A E = C D - D E (G i v e n A B = C D)$

$\Rightarrow B E = C E$

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