IIT-JEE 1980Question 1
$ABC$ is a triangle with $\angle B$ greater than $\angle C$. $D$ and $E$ are points on $BC$ such that $AD$ is perpendicular to $BC$ and $AE$ is the bisector of angle $A$. Complete the relation $$\angle DAE=\frac12\left[(\ldots)-\angle C\right]$$
Points and Straight Lines
Answer$\angle B$
Solution
$\angle BAD=90^\circ-B$, $\angle BAE=\frac A2=90^\circ-\frac{B+C}2$, so $\angle DAE=\angle BAE-\angle BAD=\frac12(B-C)$.