Question

In given figure $AB$ is a diameter of a circle. $CD$ is a chord equal to the radius of the circle. $AC$ and $BD$ when extended intersect at a point $E$. Prove that $\angle$AEB= ${60}^{\circ }$.

Answer 1

Join $OC$ and $OD$. and $CB$

We have, $OA=OB=OC=OD=CD$.
Thus, $\Delta COD$ is an equilateral triangle.
$\therefore \angle COD={60}^{\circ }$
The angle subtended at the centre of a circle by a chord has measure twice that of an angle subtended on the perimeter by the same chord.
Thus, $\angle CBD=\frac{{60}^{\circ }}{2}={30}^{\circ }$
Since AB is the diameter and C is a point on the semi circle, we have, $\angle ACB={90}^{\circ }$
$\Rightarrow \angle BCE={90}^{\circ }[\because \angle ACB$ and $\angle BCE$ are linear pair]
Now consider the triangle $\Delta CBE.$
By angle sum property, we have
$\angle BCE+\angle CBE+\angle CEB={180}^{\circ }$
$\Rightarrow {90}^{\circ }+{30}^{\circ }+\angle CEB={180}^{\circ }$
$\Rightarrow \angle CEB={180}^{\circ }-{90}^{\circ }-{30}^{\circ }$
$\Rightarrow \angle CEB={180}^{\circ }-{120}^{\circ }$
$\Rightarrow \angle CEB={60}^{\circ }$
$\Rightarrow \angle AEB={60}^{\circ }$