# Prove that $AA_1, BB_1,CC_1$ are concurrent at $O$ and $M,O,G$ are collinear.

Let $$\Delta ABC$$ and any point $$M$$ in the triangle. $$A_1,B_1$$ and $$C_1$$ be symmetric points of $$M$$ through the midpoints of $$BC, CA, AB$$. Prove that $$AA_1, BB_1,CC_1$$ are concurrent at $$O$$ and $$M,O,G$$ are collinear.

I proved that $$BCB_1C_1$$ is parallelogram then it has done but i am studying about vector then i want to solve it by vector. Otherwise i thought we will assume $$AA_1\cap BB_1=O$$ then $$C_1,O,C$$ are collinear. Indeed i tried to prove $$\overrightarrow{CO}=k\overrightarrow{CC_1}$$ but failed. Help me

The solution by vectors.

Let $$\overrightarrow{MA}=\vec{a}$$, $$\overrightarrow{MB}=\vec{b}$$, $$\overrightarrow{MC}=\vec{c}$$, $$A_2$$, $$B_2$$ and $$C_2$$ be midpoints of $$BC$$, $$AC$$ and $$AB$$ respectively

and let $$M_1$$ be a point such that $$\overrightarrow{MA}=\overrightarrow{A_1M_1}.$$

Thus, $$\overrightarrow{BM}=\overrightarrow{BC_2}+\overrightarrow{C_2M}=\overrightarrow{C_2A}+\overrightarrow{C_1C_2}=\overrightarrow{C_1A}.$$ By the same way we obtain: $$\overrightarrow{MB}=\overrightarrow{AC_1}=\overrightarrow{B_1M_1}=\overrightarrow{CA_1}=\vec{b},$$ $$\overrightarrow{MA}=\overrightarrow{BC_1}=\overrightarrow{A_1M_1}=\overrightarrow{CB_1}=\vec{a}$$ and $$\overrightarrow{MC}=\overrightarrow{AB_1}=\overrightarrow{C_1M_1}=\overrightarrow{BA_1}=\vec{c}.$$ Now, let $$AA_1\cap BB_1=\{O\}$$.

Thus, $$\overrightarrow{CO}=\frac{1}{2}(\overrightarrow{CB}+\overrightarrow{CB_1})=\frac{1}{2}(-\vec{c}+\vec{b}+\vec{a})=\frac{1}{2}(\overrightarrow{B_1C_1}+\overrightarrow{BC_1})=\overrightarrow{OC_1},$$ which says that $$O$$ is a common point of $$AA_1$$, $$BB_1$$ and $$CC_1$$.

Now, $$\overrightarrow{MG}=\overrightarrow{MB}+\overrightarrow{BG}=\vec{b}+\frac{2}{3}\cdot\frac{1}{2}(\overrightarrow{BC}+\overrightarrow{BA})=$$ $$=\vec{b}+\frac{1}{3}(-\vec{b}+\vec{c}-\vec{b}+\vec{a})=\frac{1}{3}(\vec{a}+\vec{b}+\vec{c}).$$ Also, $$\overrightarrow{MO}=\frac{1}{2}\left(\overrightarrow{MB}+\overrightarrow{MB_1}\right)=\frac{1}{2}(\vec{b}+\vec{c}+\vec{a})$$ and we got that $$M$$, $$O$$ and $$G$$ are placed on the same line.

Since $$BMCA_1$$, $$BMAC_1$$ and $$MCB_1A$$ are parallelograms, we obtain that $$A_1C_1AC$$, $$A_1B_1AB$$ and $$CB_1C_1B$$ are parallelograms, which gives that $$AA_1$$, $$BB_1$$ and $$CC_1$$ are concurrent because any two of them they are diagonals of one parallelogram from three last parallelograms.

Now, we can see the picture as a projection of a parallelepiped $$A_1BMCM_1C_1AB_1,$$ where $$O$$ is a midpoint of $$MM_1.$$

$$AA_1$$, $$BB_1$$, $$CC_1$$ and $$MM_1$$ they are diagonals of the parallelepiped.

Now, easy to get also the second part of the problem because $$MM_1\cap(ABC)=\{G\}$$

and we got also that $$MG=\frac{1}{3}MM_1$$ and $$MO=\frac{1}{2}MM_1$$.

• Thank you Michael but i want a solution by vector. Aug 5, 2019 at 14:53