Prove that $BC$, $B_1C_1$, $B_2C_2$ are concurrent. 
Consider altitude $AH$ of $\Delta ABC$. $B_1$ and $B_2$ are points on side $AB$ such that $HB_1 \perp AB$ and $HB_2 \parallel AC$. $C_1$ and $C_2$ are points on side $AC$ such that $HC_1 \perp AC$ and $HC_2 \parallel AB$. Prove that $BC$, $B_1C_1$,  $B_2C_2$ are concurrent.


I tried letting $B_1C_1 \cap B_2C_2 = \{A'\}$ and tried to prove that $\widehat{AA'B} = \widehat{AA'C}$ but I don't see the light at the end of the tunnel.

First, notice that triangles $BB_2H$ and $BAC$ are similar. This gives the following:
$$\frac{BB_2}{c}=\frac{c\cos\beta}{a}$$
$$BB_2=\frac{c^2\cos\beta}{a}$$
$$\frac{BB_2}{B_2A}=\frac{BB_2}{c-BB_2}=\frac{c\cos\beta}{a-c\cos\beta}\tag{1}$$
In the same way you can show that:
$$\frac{C_2C}{AC_2}=\frac{b\cos\gamma}{a-b\cos\gamma}\tag{2}$$
From (1) and (2)
$$\frac{BB_2}{B_2A}\cdot\frac{AC_2}{C_2C}=\frac{c\cos\beta}{a-c\cos\beta}\cdot\frac{a-b\cos\gamma}{b\cos\gamma}\tag{3}$$
If you use the fact that:
$$\frac ac=\frac{\sin\alpha}{\sin\gamma},\ \ \sin\alpha=\sin(\beta+\gamma)$$
...you can actually show (in a pretty trivial way) that (3) leads to:
$$\frac{BB_2}{B_2A}\cdot\frac{AC_2}{C_2C}=\frac{\tan^2\gamma}{\tan^2\beta}\tag{4}$$
On the other side you can easily show that:
$$BB_1=c\cos^2\beta, \ B_1A=c-BB_1=c\sin^2\beta$$ 
$$C_1C=b\cos^2\gamma, \ AC_1=b-C_1C=b\sin^2\beta$$ 
...which leads to:
$$\frac{BB_1}{B_1A}\cdot\frac{AC_1}{C_1C}=\frac{\tan^2\gamma}{\tan^2\beta}\tag{5}$$
By comparing (4) and (5):
$$\frac{BB_1}{B_1A}\cdot\frac{AC_1}{C_1C}=\frac{BB_2}{B_2A}\cdot\frac{AC_2}{C_2C}\tag{6}$$
Now introduce points $A'=BC\cap B_1C_1$, $A''=BC\cap B_2C_2$.
By Menelaus, you can write (6) as:
$$\frac{A'B}{CA'}=\frac{A''B}{CA''}$$
...which simply means that points $A'$ and $A''$ are identical. Consequentially, lines $BC$, $B_1C_1$ and $B_2C_2$ are concurrent.
