Question

Let $ABC$ be a triangle with $\angle A={90}^o$ and $AB=AC.$ Let $D$ and $E$ be points on the segment $BC$ such that $BD:DE:EC=1:2:\sqrt{3}$. Find $\angle DAE$.

• A. ${25}^o$
• B. ${35}^o$
• C. ${45}^o$
• D. ${55}^o$

Answer 1

The correct option is C ${45}^o$

Rotating the conguraiton about $\angle A$ by 90 degree, the point B goes to the point C. Let P denote the image of the point D under this rotation.
$\therefore CP=BD$ and $\angle ACP=\angle ABC={45}^o$
Therefore $ECP$ is a right-angled triangle with $CE:CP=\sqrt{3}:1$
Hence $PE=ED$
It follows that ADEP is a kite with AP = AD and PE = ED
Therefore AE is the angular bisector of $\angle PAD$. This implies that $\angle DAE=\angle PAD/2={45}^o$.