Let $A_1, B_1, C_1$ be the midpoints of the sides $BC, CA, AB$ of triangle $ABC$ respectively. Let $M$ be an arbitrary point in the plane of the triangle. Prove that $\vec{MA_1} + \vec{MB_1} + \vec{MC_1} = \vec{MA} + \vec{MB} + \vec{MC}$.

1. Problem Description

Let $A_1, B_1, C_1$ be the midpoints of the sides $BC, CA, AB$ of triangle $ABC$ respectively. Let $M$ be an arbitrary point in the plane of the triangle. Prove that $\vec{MA_1} + \vec{MB_1} + \vec{MC_1} = \vec{MA} + \vec{MB} + \vec{MC}$.

2. Solution Steps

Since $A_1$ is the midpoint of $BC$, we have

$\vec{MA_1} = \frac{1}{2}(\vec{MB} + \vec{MC})$.

Similarly, since $B_1$ is the midpoint of $CA$, we have

$\vec{MB_1} = \frac{1}{2}(\vec{MC} + \vec{MA})$.

And since $C_1$ is the midpoint of $AB$, we have

$\vec{MC_1} = \frac{1}{2}(\vec{MA} + \vec{MB})$.

Adding these three equations, we get:

$\vec{MA_1} + \vec{MB_1} + \vec{MC_1} = \frac{1}{2}(\vec{MB} + \vec{MC}) + \frac{1}{2}(\vec{MC} + \vec{MA}) + \frac{1}{2}(\vec{MA} + \vec{MB})$

$\vec{MA_1} + \vec{MB_1} + \vec{MC_1} = \frac{1}{2}\vec{MB} + \frac{1}{2}\vec{MC} + \frac{1}{2}\vec{MC} + \frac{1}{2}\vec{MA} + \frac{1}{2}\vec{MA} + \frac{1}{2}\vec{MB}$

$\vec{MA_1} + \vec{MB_1} + \vec{MC_1} = \vec{MA} + \vec{MB} + \vec{MC}$.

3. Final Answer

$\vec{MA_1} + \vec{MB_1} + \vec{MC_1} = \vec{MA} + \vec{MB} + \vec{MC}$.