# 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

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:

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