If the hypotenuse is $4$ times the height from $A$, prove that one of the angles is $15^\circ$

Question

In a right triangle (with $\angle CAB = 90^\circ$), suppose $|BC| = 4|AD|$ with $AD$ being the height from $A$ to $BC$. Prove that $\angle BCA$ is $15^\circ$.

I had a similar problem but with $22.5^\circ$. I thought it would be similar, but I found nothing I could do. Could you help me please?

Answer 1

Note that: by the Pitagoras Theorem $(\overline{CA})^2+(\overline{AB})^2=(\overline{BC})^2$, and by the area of triangle $\frac{(\overline{BC})^2}{4}=(\overline{CA})\cdot(\overline{AB})$

Now if $\measuredangle BCA=\alpha$ then: $$(\sin{\alpha}+\cos{\alpha})^2=\left(\frac{\overline{AB}}{\overline{BC}}+\frac{\overline{AC}}{\overline{BC}}\right)^2$$ then: $$1+\sin{2\alpha}=\frac{3}{2}$$ then $$2\alpha=30^{\circ}\Longrightarrow \alpha=15^{\circ}$$

Answer 2

Hint. If $\alpha=\angle BCA$ then looking at the similar triangles $\triangle ABC,\,\triangle DAC,\,\triangle DBA$ gives $$\tan\alpha+\cot\alpha=4\ ,$$ which can be solved by using standard trig formulae. See if you can fill in the details.

Answer 3

Hint: We have $$a^2+b^2=c^2 $$ and (calculate area in two different ways) $$ ab=\frac{c^2}{4}. $$

So: $$a^2+b^2=4ab$$
Dividing by $ab$ both sides and setting $x=\frac{a}{b}$, this gives us: $$ x+\frac{1}{x}=4.$$

Once we have $x$, that is we have $\frac{a}{b}$, invert the relevant trigonometric function to get the needed angle.