In triangle $ABC$ if $AC=BD$ , $\angle ABC=\angle CAD=18^{\circ}$, $\angle BAD=?$.

Question

In triangle $ABC$ $AC=BD$ , $\angle ABC=\angle CAD=18^{\circ}$.

Find the angle $\angle BAD$.

I tried to solve it in basic way but it doesn't seem to work:

How would you solve this? Any method is okay.

Answer 1

Note that $\cos(36^\circ)=\cos\left(\frac{\pi}{5}\right)=\frac{\sqrt 5+1}{4}$, use $\cos(36^\circ)=1-2\sin^2(18^\circ)$, we can find the value for

$$\sin(18^\circ)=\frac{\sqrt 5-1}{4}\tag{1}$$

Now we are done for the preparation work.

$$\frac{\sin x}{BD}=\frac{\sin (18^\circ)}{AD},~~\frac{\sin(18^\circ+x)}{AC}=\frac{\sin C}{AD},~~ C=180^\circ-x-18^\circ-18^\circ$$

Combine above equations, we get,

$$\sin x\sin(36^\circ+x)=\sin (18^\circ)\sin(18^\circ+x)$$

For LHS, use: $\sin A\sin B=\frac{\cos(A-B)-\cos(A+B)}{2}$, we get,

$$\cos (36^\circ)-\cos(36^\circ+2x)=2\sin (18^\circ)\sin(18^\circ+x)$$

Define $\theta=18^\circ+x$,

$$\cos (36^\circ)-\cos(2\theta)=2\sin (18^\circ)\sin(\theta)$$

Use double angle formula for LHS, we get,

$$-2\sin^2 (18^\circ)+2\sin^2(\theta)=2\sin (18^\circ)\sin(\theta)$$

Solve the quadratic equation, and plug in the value from $(1)$

$$\sin\theta=\frac{1+\sqrt5}{2}\sin(18^\circ)=\frac{1+\sqrt5}{2}\cdot \frac{\sqrt 5-1}{4}=\frac{1}2$$

Therefore we get:

$$\theta=30^\circ\Rightarrow x=12^\circ$$

Answer 2

Let $AD=BC=x, CD=y, \angle BAD= \alpha $

then

$\angle ADC= \alpha+ 18^{\circ}$

$\triangle ACD \sim \triangle BAD$ (A.A.A. similarity)

$\frac{CD}{AD}=\frac{AD}{BD}$

$\frac{y}{x}=\frac{x}{x+y}, x^2-yx-y^2=0$

By solving equation in terms of $y$

$x=\frac{(1\pm \sqrt5)y}{2}$ then $x=\frac{(1+ \sqrt5)y}{2}$

Applying Sine Theorem in $\triangle ACD$

$\frac{x}{y}=\frac {\sin (\alpha +18^{\circ})}{\sin 18}=\frac{1+ \sqrt5}{2}$

Since $\sin 18 = \frac{\sqrt5 - 1}{4}$

$\sin (\alpha +18^{\circ}) = \frac{\sqrt5 - 1}{4} \times \frac{1+ \sqrt5}{2}=\frac{1}{2}=\sin 30^{\circ} = \sin 150^{\circ} $

$\alpha+18^{\circ}=30^{\circ}, \alpha=12$

$\alpha+18^{\circ}=150^{\circ}, \alpha=132$

Answer 3

Note that $\triangle ABC \sim \triangle DAC$ because $\angle CBA = \angle DAC$ and a common angle at the vertex $C$. It follows that $$\frac{BC}{CA} = \frac{CA}{CD}.$$ Let $E$ be the point such that $DE=CE=AC$ which lies on the same side of the line $BC$ as $A$. Then $$\frac{BC}{CE} = \frac{BC}{CA} = \frac{CA}{CD} = \frac{CE}{CD}.$$ We conclude by SAS that $\triangle BCE \sim \triangle ECD$. Denote $\angle DEC = \vartheta$. Then $\angle CBE = \vartheta$. Note that $BD = DE$, hence $\angle BED = \angle DBE = \vartheta$. Therefore $\angle CDE = \angle DBE + \angle BED = 2\vartheta$. Since the triangle $EDC$ is isosceles, $\angle ECD = \angle CDE = 2\vartheta$. Summing up internal angles of $\triangle EDC$ we obtain $5\vartheta = 180^\circ$, so $\vartheta=36^\circ$. Since $\angle CBA = 18^\circ = \frac 12 \vartheta = \frac 12 \angle CBE$, the line $BA$ bisects the angle $CBE$. But triangle $CBE$ is isosceles (as it is similar to $EDC$), hence $BA$ is the perpendicular bisector of $CE$. Therefore $AE=AC$ from which we see that the triangle $ACE$ is equilateral. Hence $\angle BAC = \frac 12 \angle EAC = 30^\circ$ and finally $$\angle BAD = 30^\circ - 18^\circ = 12^\circ.$$