Question

Let $ABC$ be a right triangle with length of side $AB=3$ and hypotenuse $AC=5$. If $D$ is a point on $BC$ such that $\frac{BD}{DC}=\frac{AB}{AC}$, then $AD$ is equal to:

• A. $\frac{4\sqrt{3}}{3}$
• B. $\frac{3\sqrt{5}}{2}$
• C. $\frac{4\sqrt{5}}{3}$
• D. $\frac{5\sqrt{3}}{4}$

Answer 1

The correct option is B $\frac{3\sqrt{5}}{2}$

Given: $\frac{BD}{DC}=\frac{AB}{AC}$
$\Rightarrow AD$ is the angle bisector.
$\therefore AD=\frac{2bc}{b+c}\cos \frac{A}{2}$
$=\frac{2\cdot 5\cdot 3}{8}\sqrt{\frac{1+\cos A}{2}}=\frac{15}{4}\sqrt{\frac{1+\frac{3}{5}}{2}}$
$=\frac{3\sqrt{5}}{2}$