Question

In the given figure, ray $AD$ is the bisector of $\angle CAB$ and the measure of $\angle ACD$ is $y$. What must be the measure of $\angle ADC$ in order for line $AB$ to be parallel to line $CD$.

• A. ${90}^o-y$
• B. ${90}^o-y/2$
• C. ${180}^o-y/2$
• D. ${90}^o+y$

Answer 1

The correct option is B ${90}^o-y/2$

For $AB$ to be parallel to $CD$
$\angle BAC+y={180}^o$ $\Rightarrow$ $\angle BAC={180}^o-y$
Since, $AD$ bisects $\angle BAC$, $\angle BAD={90}^o-y/2$
Also, $AB\parallel CD$
$\Rightarrow$ $\angle ADC=\angle BAD={90}^o-y/2$. (alt. $\angle s$)