Question

ABCD is a cyclic quadrilateral in which AD is the diameter and $\angle BCD={125}^o$, then the measure of $\angle ADB$ is

• A. ${35}^o$
• B. ${45}^o$
• C. ${55}^o$
• D. ${65}^o$

Answer 1

The correct option is A ${35}^o$

In given figure ABCD is a cyclic quadrilateral.

We know that, sum of the opposite angles in a cyclic quadrilateral is 180${}^{\circ }$

So, $\angle$BCD + $\angle$BAD = 180${}^{\circ }$

$\Rightarrow$ 125${}^{\circ }$ + $\angle$BAD = 180${}^{\circ }$

$\Rightarrow$ $\angle$BAD = 180${}^{\circ }$ - 125${}^{\circ }$

$\therefore$ $\angle$BAD = 55${}^{\circ }$ ---- (1)

Since, $\angle$ABD is an angle in semi-circle.

$\therefore$ $\angle$ABD = 90${}^{\circ }$ --- (2)

In $△$ABD,

$\angle$ADB + $\angle$ABD + $\angle$BAD = 180${}^{\circ }$

$\angle$ADB + 90${}^{\circ }$ + 55${}^{\circ }$ = 180${}^{\circ }$ (From 1 and 2)

$\therefore$ $\angle$ADB = 35${}^{\circ }$