If $ \bigtriangleup ABC$: $\angle CAB = \frac{\pi}{2}$, with height $AD$ and median $AK$. Prove $\angle BAD = \angle BCA = \angle KAC.$

Question

If $\triangle ABC$ is a triangle and $\angle CAB = \frac{\pi}{2}$, with height $AD$ and median $AK$; suppose that $D$ is between $B$ and $K$.

1. Prove that $\angle BAD = \angle BCA = \angle KAC$.
2. Then, prove that $\angle BCA= \frac\pi 8$ if $|AD|=|DK|$.
3. Conclude that $$\sin\frac \pi 8=\frac{\sqrt{2-\sqrt 2}}{2};\quad \cos\frac \pi 8=\frac{\sqrt{2+\sqrt 2}}{2};\quad \operatorname{tg}\frac \pi 8= \sqrt 2 - 1$$

I already achieved the draw, but I don't really know how to start.

I know that $\bigtriangleup DBA \sim \bigtriangleup DAC$

Answer 1

Consider the circumcircle of $\triangle ABC$. Since $\angle A=\frac{\pi}{2}$, it subtends the diameter, thus $K$ is the circumcenter and $$KA=KB=KC\tag{1}$$

1. Since $\triangle KCA$ is isosceles, $\angle KCA=\angle KAC$.
   In $\triangle ABD$ $\ \ \angle D=\frac{\pi}{2}$, thus $\angle BAD=\frac{\pi}{2}-\angle ABD$, but $\frac{\pi}{2}-\angle ABC=\angle ACB$, thus $\angle BAD=\angle ACB=\angle KAC$, QED.
2. In $\triangle ADK$ $\ \ \angle D=\frac{\pi}{2}$, thus $|AD|=|DK|$ $\Rightarrow$ $\angle A=\angle K=\frac{\pi-\angle D}{2}=\frac{\pi}{4}$.
   Since $\frac{\pi}{4}=\angle AKD=\angle KAC+\angle KCA$ and $\angle KAC=\angle KCA$, thus $\angle ACK=\frac{\pi}{8}$, QED.
3. In $\triangle ADC$ $\ \ \angle D=\frac{\pi}{2}$ and $AK=KC=AD\sqrt{2}$ thus $$\tan \frac{\pi}{8}=\frac{AD}{DK+KC}=\frac{AD}{AD+AD\sqrt{2}}= \frac{1}{1+\sqrt{2}}=\sqrt{2}-1,$$ the other functions of $\frac{\pi}{8}$ are done by using $$\frac{1}{\cos^2\theta}=1+\operatorname{tg}^2\theta,\quad \frac{1}{\sin^2\theta}=1+\operatorname{ctg}^2\theta.$$

Answer 2

Let $D$ be placed between $K$ and $B$.

Thus, since $AK$ is a median, we obtain $$AK=CK=KB,$$ which gives $$\measuredangle BAD=90^{\circ}-\measuredangle ABC=\measuredangle BCA=\measuredangle KAC.$$

Answer 3

1. Since you have figured out that $\triangle DBA \sim \triangle DAC$, use the property that corresponding angles of similar triangles are equal. Also, notice that $AK=KC$, hence $\triangle KAC$ is isosceles.

2. If $AD=DK$, we have $\angle DKA=\angle KAD=45°\implies\angle AKC=135°$. Thus, $\triangle KAC$ being isosceles, we have $\angle BCA=22.5°=\frac{π}{8}$.

3. We have $AK=KC=\frac{a}{2}\implies AD=DK=\frac{a}{2\sqrt 2}$. In $\triangle ADC$, $$\tan\angle DCA=\tan\frac{π}{8}=\sqrt 2-1$$