Let $A_1, B_1, C_1$ be the midpoints of 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 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

We are given that $A_1$, $B_1$, and $C_1$ are midpoints of $BC$, $CA$, and $AB$, respectively. Therefore, we have:

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

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

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

Now, let's sum these three equations:

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

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

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

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

Thus, we have proven that $\vec{MA_1} + \vec{MB_1} + \vec{MC_1} = \vec{MA} + \vec{MB} + \vec{MC}$.

3. Final Answer

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