In $\triangle ABC$, if the altitude, median, and angle bisector at $A$ quadrisect $\angle A$, then $\angle A=90^\circ$ and $\angle C=22.5^\circ$ The problem and answer are from a book.

Is there a $\triangle ABC$ such that the altitude from $A$, the bisector of $\angle BAC$ and the median from $A$ divide $\angle BAC$ into four equal parts?
The answer is:
$ABC$ is a right triangle with $\angle A=90^\circ$ and $\angle C=22.5^\circ$.

I don't know how I should start to solve this problem.
 A: This is not difficult.
Let the altitude be $AE$, the bisector $AD$ and the median $AM$.
Let $\angle A=4\alpha$. Since $\angle BEA=90^\circ$, we have $\angle B=90-\alpha$. Hence $\angle C=90-3\alpha$ (because the angles in $ABC$ add to $180^\circ$).

From triangle $MAC$ the sine rule gives $MC/MA=\sin\alpha/\sin(90^\circ-3\alpha)=\sin\alpha/\cos3\alpha$. Similarly, from triangle $MAB$ we get $MB/MA=\sin3\alpha/\cos\alpha$. But $MB=MC$, so $\sin3\alpha\cos3\alpha=\sin\alpha\cos\alpha$. Hence $\sin6\alpha=\sin2\alpha$, so $6\alpha+2\alpha=180^\circ$ and hence $\alpha=22.5^\circ$ and so $\angle BAC=90^\circ$.
A: Well the angle bisector cuts $\angle BAC$ into two equal parts, so the median and the altitude must each cut those two angles into equal parts.  If we label point where the altitude intersects $BC$ as $J$, the point where the angle bisector as $K$, and the median as $M$, we know that $K$ is colinearly between $J$ and $M$.  As it is arbitrary which endpoint is $B$ and which is $C$ we can assume that points lie in order as $B,J,K,M$
So we have a figure a big triangle $\triangle ABC$ divided into four smaller triangles.
$\angle BAJ \cong \angle JAK \cong \angle KAM \cong \angle MAC$.
$\angle BJA \cong \angle KJA$ are both right angles.
So $\triangle BJA \cong \triangle KJA$.
If we let $m\angle BAJ = m\angle JAK = m\angle KAM = m \angle MAC= X$ then we can conclude:
$m\angle ABJ = 90 -X$ and $m\angle BCA = 180-(90-X)-4x = 90-3X$.
Now if we look at the line $BC$ and use trig identities we know.
$\frac {BJ}{AJ} = \tan X$.
$\frac {KJ}{AJ} = \tan X$. (And $BJ=JK$)$
$\frac {MJ}{AJ} = \tan 2X$.
And $\frac {CJ}{AJ} = \tan 3X$.
And heres the lynchpin:  $M$ is the midpoint of $BC$ so $MC = MB$.
Now $MC = CJ - MJ = AJ(\tan 3X - \tan 2X)$ and $MB = BJ + MJ = AJ(\tan X + \tan 2X)$.
So we have $\tan 3X - \tan 2X = \tan X + \tan 2X$
A: Here is a possible solution which is not using trigonometric functions, but rather only similarities and the angle bisector theorem (ABT). 

We denote by $H,J,M$ the points on $BC$ which are respectively on the height, angle bisector, and median of $A$. Let $K$ be the reflection of $J$ w.r.t. the median $AM$. It lies on the side $AC$. Let $a,b,c$ be the side lengths as usual. Let the angle in $A$ have the measure $4x$. Then we can quickly chase the following angles in the picture: 
$A$ divide the angle in $A$ in four equal pieces">
Now we compute the lenghts in the picture.


*

*From the ABT in $\Delta AJC$ we get the length of $JM$ as shown in the picture, $\frac a2\cdot \frac cb$. So we also have $MK$. 

*$BJ=BM-JM$ leads to $BJ=\frac a2\cdot \frac {b-c}b$.

*The rest of the segment $BC=a$ is then $JC=\frac a2\cdot {b+c}b$.


We write now the ABT in the triangle $\Delta ABC$, and the similarity $\Delta ABC\sim \Delta MKC$ to get the relations:
$$
\begin{aligned}
\frac cb=\frac{b-c}{b+c}\ ,\\
\frac ab=\frac{b-c}{a/2}\ .
\end{aligned}
$$
The two realtions are $b^2-bc=c^2+bc$ and $a^2=2b^2-2bc$. It follows immediately $a^2=b^2+(b^2-2bc)=b^2+c^2$, so the angle in $A$ is $4x=90^\circ$, and all angles can be explicitly identified.

To be pedant, we have to check if the obtained triangle satisfies the condition, yes, this is the case, the height $AH$ is the angle bisector of $\widehat{BAJ}$, 
and because of $MA=MC$ we have $\widehat{MAC}=\widehat{MCA}=22.5^\circ$, so $AM$ bisects $\widehat{JAC}$.
