Question

In an isosceles triangle $ABC,AB=AC,\angle BAC={108}^o$ and $AD$ trisects $\angle BAC$ and $BD>DC$. The ratio $\frac{BD}{DC}$ is

• A. $\frac{2}{3}$
• B. $\frac{\sqrt{5}+1}{2}$
• C. $\sqrt{5}-1$
• D. $2$

Answer 1

The correct option is A $\frac{2}{3}$

Given,

AB=AC, $\angle BAC={108}^{\circ }$

So $\angle ABC=\angle ACB=\frac{180-\angle BA}{2}=\frac{180-108}{2}={36}^{\circ }$

In $\Delta$ABD

$\frac{BD}{sin36}=\frac{AB}{sin108}-\left(1\right)$

In $\Delta$ ACD

$\frac{CD}{sin72}=\frac{AC}{sin72}⟹CD=AC⟹CD=AC=AB$

eqs.(1)

$⟹\frac{BD}{sin36}=\frac{CD}{sin108}⟹\frac{BD}{CD}=\frac{sin36}{sin108}$

$\frac{BD}{CD}=\frac{2}{3}$