In $\Delta ABC$, $AC=\sqrt{3}AB$ and $BC=2$. D is a point inside $\Delta ABC$ such that $\angle BDC=90^\circ, \angle DAC=18^\circ$ and $BD=1$

Question

Question: In $\Delta ABC$, $AC=\sqrt{3}AB$ and $BC=2$.
D is a point inside $\Delta ABC$ such that $\angle BDC=90^\circ, \angle DAC=18^\circ$ and $BD=1$, find $\angle BAD$.

I can solve this problem using trigonometry and I get $\angle BAD = 48^\circ$. But I wonder if this can be done without the use of trigonometry. I have tried many different ways to construct isosceles triangles, adding new lines etc but still cannot solve the problem.

Answer 1

As in the figure above, draw $AE$ in such a way that $AE \perp AC$, and $AE\cong AB$.

1. By SAS criterion $\triangle AEC \sim \triangle BDC$.
2. $\measuredangle ECB = \measuredangle ACD$, since both are equal to $30^\circ - \measuredangle DCE.$
3. Consequently (SAS criterion) $\triangle BEC \sim \triangle ADC$, and, in particular, $\measuredangle BEC = 18^\circ$.
4. Since $\measuredangle AEC = 60^\circ$ and $\triangle AEB$ is isosceles, we get $\measuredangle EAB = 24^\circ$.
5. Thus $\measuredangle BAC = 66^\circ$, and from that $$\boxed{\measuredangle BAD = 48^\circ}.$$

Answer 2

Let the angle we need to find be $\theta$

With elementary trig u can find out that $\angle DBC=60$ and $\angle DCB=30$

Let's create the $\triangle BAM$ such as $\angle BAM=90, \angle ABM=60, \angle ADM=30$

Let the side $AB=x$ by elementary trig as before we can say $BM=2x, AM=\sqrt{3}x$ And from the data given in the question $AC=\sqrt{3}x$

Look at $\triangle ABD, \triangle MBC$, By simple angle chasing we can find out $\angle ABD = \angle MBC$

also $\frac{AB}{BD}=\frac{x}{1}$ and $\frac{BM}{BC}=\frac{2x}{2}=\frac{x}{1}$

Therefore the $\triangle ABD$ and $\triangle MBC$ are similar. Hence $\angle BAD=\angle BMC=\theta$

Look closely at the $\triangle CAM$ it is isosceles. Therefore $\angle AMC=\angle MCA=30+\theta$ and $\angle CAM=120-2\theta$

Finally as $\angle BAM=90$,

$\theta+18+120-2\theta = 90$ and $\boxed{\theta=48^o}$

Thank u for sharing such a creative problem with us !