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Two chords AB and CD intersect each other at the point E as shown in the given figure. Prove that $\Delta AED \sim \Delta CEB$.
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Solution (a) In $\Delta AED$ and $\Delta CEB$,
$\angle DAB = \angle DCB$ (angles in same segment of a circle) (1 Mark)
$\angle AED = \angle CEB$ (vertically opposite angles) (1 Mark)
$\Delta AED \sim \Delta CEB$ (by AA similarity)
$\angle DAB = \angle DCB$ (angles in same segment of a circle) (1 Mark)
$\angle AED = \angle CEB$ (vertically opposite angles) (1 Mark)
$\Delta AED \sim \Delta CEB$ (by AA similarity)