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

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Solution

In $△ O M E$ and $△ O N E$ ,
$O M = O N$ because equal chords of a circle are equidistant from the centre
$O E = O E$ (common)
$∴ △ O M E ≅ △ O N E$ by R.H.S
$∴ M E = N E$ by C.P.C.T
$\Rightarrow A M + M E = D N + N E$
$∵ A M = D N = \frac{1}{2} A B = \frac{1}{2} C D$
$\Rightarrow A E = D E$
$\Rightarrow A B - A E = C D - D E$ because $A B = C D$
$\Rightarrow B E = C E$
$∴ A E = D E$ and $B E = C E$ .
Hence proved

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

In △OME and △ONE,OM=ON because equal chords of a circle are equidistant from the centreOE=OE(common)∴△OME≅△ONE by R.H.S∴ME=NE by C.P.C.T⇒AM+ME=DN+NE∵AM=DN=12AB=12CD⇒AE=DE⇒AB−AE=CD−DE because AB=CD⇒BE=CE∴AE=DE and BE=CE.Hence proved