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 $\angle ACD + \angle BED$.
$(270^{\circ})$ Join $BC$.
Since the angle in a semicircle is $90^{\circ}$,we have
$\angle ACB = 90^{\circ}$.
As $ACDEB$ is a cyclic quadrilateral (all vertices lie on the circle),the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle BCD + \angle BED = 180^{\circ}$.
Now,adding $\angle ACB$ to both sides,we get:
$(\angle BCD + \angle ACB) + \angle BED = 180^{\circ} + \angle ACB$
Since $\angle BCD + \angle ACB = \angle ACD$,we have:
$\angle ACD + \angle BED = 180^{\circ} + 90^{\circ} = 270^{\circ}$.
Since the angle in a semicircle is $90^{\circ}$,we have
$\angle ACB = 90^{\circ}$.
As $ACDEB$ is a cyclic quadrilateral (all vertices lie on the circle),the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle BCD + \angle BED = 180^{\circ}$.
Now,adding $\angle ACB$ to both sides,we get:
$(\angle BCD + \angle ACB) + \angle BED = 180^{\circ} + \angle ACB$
Since $\angle BCD + \angle ACB = \angle ACD$,we have:
$\angle ACD + \angle BED = 180^{\circ} + 90^{\circ} = 270^{\circ}$.
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