Question

Question 19
In the figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of $\angle ACD+\angle BED$.

Answer 1

In the given figure, join AE.
Since, A, C, D and E are four points on a circle, then ACDE is a cyclic quadrilateral.
$\angle ACD+\angle AED={180}^{\circ }$...........(i) [sum of opposite angles in a cyclic quadrilateral is ${180}^{\circ }$]
Now, $\angle AEB={90}^{\circ }$..............(ii) [The diameter subtends a right angle to the circle.]

On adding Eqs. (i) and (ii), we get
$(\angle ACD+\angle AED)+\angle AEB={180}^{\circ }+{90}^{\circ }={270}^{\circ }$
$\Rightarrow \angle ACD+\angle BED={270}^{\circ }$

OK