Question

If two equal chords of a circle intersect, then prove that their segments are equal.
$\left[3Marks\right]$

Answer 1

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 $\Delta ONE$
$\angle M=\angle N={90}^{\circ }$

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

$OE=OE$ (Common)

$\therefore$ $\Delta OME\cong \Delta ONE$ (R.H.S)
$\left[2Marks\right]$

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

$\Rightarrow AM+ME=DN+NE(\because AM=DN=\frac{1}{2}AB=\frac{1}{2}CD)$

$\Rightarrow AE=DE$

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

$\Rightarrow BE=CE$
$\left[1Mark\right]$