Question

In the figure, BD bisects $\angle ABC$ and CD bisects $\angle ACE$. The true statement is

• A. $x+y=\angle A+\angle D$
• B. $\angle D=\frac{x+y}{2}$
• C. $\angle D=\frac{1}{2}\angle A$
• D. All of the above

Answer 1

The correct option is C $\angle D=\frac{1}{2}\angle A$

$y=\frac{1}{2}(\angle B+\angle A)$
$\angle BDC=\frac{1}{2}\angle B+\frac{1}{2}\angle A-\frac{1}{2}\angle B=\frac{1}{2}\angle A$