# 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.

• Try to apply Menelaus theorem to prove that intersections of lines $BC$ and $B_1C_1$ and $BC$ and $B_2C_2$ are coincide. – richrow Mar 24 at 16:19

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.

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:

$$BB_1=c\cos^2\beta, \ B_1A=c-BB_1=c\sin^2\beta$$

$$C_1C=b\cos^2\gamma, \ AC_1=b-C_1C=b\sin^2\beta$$

$$\frac{BB_1}{B_1A}\cdot\frac{AC_1}{C_1C}=\frac{\tan^2\gamma}{\tan^2\beta}\tag{5}$$

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.