# Trigonometric Tricks in Geometry

I was digging through some old notes of questions I had done in the past and came across this one. I don’t remember who gave me this question but it was a fascinating one that combined both geometrical reasoning and the sine rule.

# Question

In the triangle $$ABC$$, $$AB \perp AC$$, $$AE=EC$$, $$AC=\sqrt{2}DE$$. Find the angle $$BDE$$.

# Solution

Spoiler

Firstly, let $$\angle ACE$$ be equal to $$\theta$$.

The triangle $$ACE$$ is isosceles as there are two equal sides $$AE$$ and $$CE$$.

$$\angle CAE$$ is equal to $$\angle ACE$$ since they are the base angles of the triangle $$ACE$$ that is isosceles. So $$\angle CAE = \theta$$.

$$\angle CAE + \angle ACE + \angle AEC = 180^\circ$$ (Angle sum of a triangle)

$\theta + \theta + \angle AEC = 180^\circ$

$\angle AEC = 180^\circ-2\theta$

Now we also know that:

$$\angle DAE = 90^\circ – \theta$$ (Complementary angles)

$$\angle EDA = 180^\circ – \alpha$$ (Angles on a straight angle)

In Triangle $$ADE$$ and using Sine Rule:

\begin{align*}\frac{\sin(180^\circ – \alpha)}{x}&=\frac{\sin(90^\circ – \theta)}{1}\\\frac{\sin \alpha}{x} &= \cos\theta\\\sin\alpha &=x\cos\theta\end{align*}

In Triangle $$ACE$$ and using Sine Rule:

\begin{align*}\frac{\sin (180^\circ – 2\theta)}{\sqrt{2}}&=\frac{\sin\theta}{x}\\\frac{\sin(2\theta)}{\sqrt{2}}&=\frac{\sin\theta}{x}\\\frac{2\sin\theta\cos\theta}{\sqrt{2}}&=\frac{\sin\theta}{x}\\\frac{2\cos\theta}{\sqrt{2}}&=\frac{1}{x}\\x\cos\theta&=\frac{\sqrt{2}}{2}\end{align*}

Equating the two results for $$x\cos\theta$$ together, we get:

\begin{align*}\sin\alpha &=\frac{\sqrt{2}}{2}\\\alpha &=45^\circ\end{align*}

[collapse]