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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.

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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.

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    $\begingroup$ Try to apply Menelaus theorem to prove that intersections of lines $BC$ and $B_1C_1$ and $BC$ and $B_2C_2$ are coincide. $\endgroup$ – richrow Mar 24 at 16:19
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Let $B_1C_1$ cuts $BC$ at $X$. Since $$AB_1\cdot AB = AH^2 = AC_1\cdot AC$$ we see that $B,C,C_1B_1$ are conyclic. Also $A,B_1,H,C_1$ are conyclic. So by the power of the point $X$ we have $$XH^2 = XB_1\cdot XC_1 = XB\cdot XC$$

So $X$ is uniqely determined by $B,C,H$.


Let $B_2C_2$ cuts $BC$ at $Y$. Now observe a homothety with the center at $Y$ which takes $B$ to $H$. Since $BB_2||HC_2$ it takes also $B_2$ to $C_2$, but then it takes line $B_2H$ to $C_2C$ (since they are parallel) and so it takes $H$ to $C$. So we have $${YB\over YH} = {YH\over YC}\implies YH^2 =YB\cdot YC$$

so $Y$ is deteremined with points $B,C$ and $H$ with the same equation, so $X=Y$ and we are done.

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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.

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