Question

In the given figure, AB is the diameter of the circle. Find the value of $\angle$ ACD

Answer 1

OB = OD (radius)

$\angle$ ODB = $\angle$ OBD [Angles opposite to equal sides of an isosceles triangle are equal]

$\angle$ ODB + $\angle$OBD + $\angle$BOD = ${180}^{\circ }$

2$\angle$ODB + ${90}^{\circ }$ = ${180}^{\circ }$

$\angle$ODB = ${45}^{\circ }$

$\angle$OBD = $\angle$ACD (Angle subtended by the common chord AD)

Therefore $\angle$ACD = ${45}^{\circ }$