Question

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

Answer 1

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 $OM\perp AB$ and $ON\perp CD$. Join OE.

Proof: In $\Delta OME$ and $\Delta ONE$

$OM=ON$ (Equal chords of a circle are equidistant from the centre)

$OE=OE$ (Common)

$\therefore$ $\Delta OME\cong \Delta ONE$ (R.H.S)

$\Rightarrow ME=NE$ (C.P.C.T)

$\Rightarrow AM+ME=DN+NE(\because AM=DN=\frac{1}{2}AB=\frac{1}{2}CD)$
[Perpendicular from centre divide the chord into two equal parts]

$\Rightarrow AE=DE$

$\Rightarrow AB-AE=CD-DE(GivenAB=CD)$

$\Rightarrow BE=CE$