One of triangle cevians is also altitude. Prove that it is angle bisector in triangle formed by foots of cevians. 
Given is a triangle $ABC$. Let $X$ be the foot of the altitude from point $A$. $P$ is a point on segment $AX$. Let $BP ∩ AC = K$ and $CP ∩ AB = L$. Prove that $\angle AXK=\angle AXL$.

I think the problem can be solved by Ceva's theorem for angles.
 A: Here is a solution using analytic geometry.  It is not too complicated, to be honest.  For a purely geometric solution, you will have to wait for a better geometer than I am.  (That geometric proof has been given at the end of this answer.)
Without loss of generality, we can suppose that $A=(0,a)$, $B=(b,0)$, $C=(c,0)$, $X=(0,0)$, and $P=(0,p)$ on the $(x,y)$-plane, where $a$, $b$, $c$, and $p$ are real numbers such that $b\neq c$ and $a\neq 0$.  If $p=a$, then $A=K=L$.  If $b=0$ or $c=0$, then $(K,L)=(P,A)$ or $(K,L)=(A,P)$.  There is nothing to prove in those cases. We suppose now that $p\neq a$, $b\neq 0$, and $c\neq 0$.  
The line $AC$ is given by the equation
$$cy=-a(x-c)$$
and the line $BK$ is given by the equation
$$by=-p(x-b)\,.$$
Therefore, the point $K$ is given by the coordinate
$$K=\left(\frac{(p-a)bc}{pc-ab},-\frac{pa(b-c)}{pc-ab}\right)\,.$$
If $pc=ab$, then $PB\parallel AC$, so $K$ is the point at infinity on the line $AC$.
Similarly, the line $AB$ is given by the equation
$$by=-a(x-b)$$
and the line $CL$ is given by the equation
$$cy=-p(x-c)\,.$$
Therefore, the point $L$ is given by the coordinates
$$L=\left(\frac{(p-a)bc}{pb-ac},-\frac{pa(c-b)}{pb-ac}\right)\,.$$
If $pb=ac$, then $PC\parallel AB$, so $L$ is the point at infinity on the line $AB$.
Thus, the slope of the line $XK$ is $$-\frac{pa(b-c)}{(p-a)bc}\,,$$ 
whilst the slope of the line $XL$ is
$$-\frac{pa(c-b)}{(p-a)bc}\,.$$
Hence, $XK$ and $XL$ have opposite slopes.  Therefore, $$\angle(XK,AX)=\angle(AX,XL)\,.$$
The equality $\angle AXK=\angle AXL$ may not be true if $X$ is not on the line segment $BC$, as seen in the picture below.


For a geometric proof (thanks to timon92, and therefore, this will become a community wiki post), for a point $T$, let $T'$ denote the image of the reflection of $T$ about the line $AX$.  Then, using signed lengths, we have
$$\frac{AL}{LB}\cdot\frac{BX}{XC'}\cdot\frac{C'K'}{K'A}=\frac{AL}{LB}\cdot\frac{BX}{(-XC)}\cdot\frac{CK}{KA}=-\left(\frac{AL}{LB}\cdot\frac{BX}{XC}\cdot\frac{CK}{KA}\right)\,.$$
By Ceva's Theorem, the Cevians $AX$, $BK$, and $CL$ of the triangle $ABC$ concur at $P$, so
$$\frac{AL}{LB}\cdot\frac{BX}{XC}\cdot\frac{CK}{KA}=1\,.$$
Thus,
$$\frac{AL}{LB}\cdot\frac{BX}{XC'}\cdot\frac{C'K'}{K'A}=-\left(\frac{AL}{LB}\cdot\frac{BX}{XC}\cdot\frac{CK}{KA}\right)=-1\,.$$
Using the converse of Menelaus's Theorem on the triangle $ABC'$ with $X\in BC'$, $K\in C'A$, and $L\in AB$, we conclude that $L$, $K'$, and $X$ are collinear.  Therefore,
$$\angle(XK,AX)=\angle(AX,XK')=\angle (AX,XL)\,.$$
A: 
Apply the sine rule to the triangle ALX and BLX
$$\frac{\sin\beta}{\sin\angle ALX} = \frac{c_1}{AX},\>\>\>\>\>\>\>\>
\frac{\sin(90-\beta)}{\sin\angle BLX} = \frac{c_2}{a_1}
$$
which leads to the expression for $\tan\beta$ below, and similarly for $\tan\gamma$
$$\tan\beta = \frac{c_1}{c_2}\frac{a_1}{AX},\>\>\>\>\>\>\>\>\tan\gamma= \frac{b_2}{b_1}\frac{a_2}{AX}$$
Take their ratio 
$$\frac{\tan\beta}{\tan\gamma} = \frac{c_1}{c_2}\frac{b_1}{b_2}\frac{a_1}{a_2}=1$$
where the Ceva's theorem is applied in the last step. Thus, $ \angle AXL = \angle AXK $.
