Question

In the given figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that$∆$ABC is an isosceles triangle.

Answer 1

In $△$AEC and $△$BEC,
AE = BE (Perpendicular drawn from the centre of the circle to the chord bisects the chord)
CE = CE (Common)
$\angle$CEA = $\angle$CEB = 90$°$
Thus, $△$AEC $\cong$ $△$BEC by SAS congurency
So, by CPCT, AC = CB
Thus, $∆$ABC is an isosceles triangle.