Show that the midpoints of two segments coincide

Question

Let $ABC$ be a triangle. $A_1,B_1$ and $C_1$ are the midpoints of $BC,AC$ and $AB$, respectively. Show that the midpoints of $CC_1$ and $A_1B_1$ coincide. Let $S_1$ and $S_2$ be the midpoints of $A_1B_1$ and $CC_1$. We have $$\vec{OS_1}=\dfrac12\left(\vec{OA_1}+\vec{OB_1}\right)=\dfrac12\left[\dfrac12\left(\vec{OB}+\vec{OC}\right)+\dfrac12\left(\vec{OA}+\vec{OC}\right)\right]\\=\dfrac14\left(\vec{OA}+\vec{OB}+2\vec{OC}\right)$$ and $$\vec{OS_2}=\dfrac12\left(\vec{OC}+\vec{OC_1}\right)=\dfrac12\left[\vec{OC}+\dfrac12\left(\vec{OA}+\vec{OB}\right)\right].$$ Are these equal?

Answer 1

Another way you could have done is using position vectors -

If positions vectors of $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ resp., then

$A_1 = \frac{\vec{b}+\vec{c}}{2}, B_1 = \frac{\vec{a}+\vec{c}}{2}, C_1 = \frac{\vec{a}+\vec{b}}{2}$

Midpoint of $A_1B_1 = \frac{1}{2} (\frac{\vec b+ \vec c}{2} + \frac{\vec a+ \vec c}{2}) = \frac{\vec a+ \vec b+ 2 \vec c}{4}$

Midpoint of $CC_1 = \frac{1}{2}(\vec c +\frac{\vec a+ \vec b}{2}) = \frac{\vec a+ \vec b+ 2 \vec c}{4}$

which shows the midpoints of both $A_1B_1$ and $CC_1$ segments coincide.

Edit: on your comment if there is a way without vector - if you are allowed to directly apply midline theorem, then here is the way but if you need to first prove $A_1B_1$ is parallel to $AB$ then see that proof here on wiki

Now given $A_1B_1 \parallel AB, \triangle ACC_1 \sim \triangle B_1CM$ (let $M$ be the point where $A_1B_1$ and $CC_1$ intersect).

So, $\frac{B_1C}{AC} = \frac{B_1M}{AC_1} = \frac{CM}{CC_1}$

As, $B_1C = \frac{AC}{2}, B_1M = \frac{AC_1}{2}, CM = \frac{CC_1}{2}$. So $M$ is the midpoint of $CC_1$.

Similarly you can show, $MA_1 = \frac{C_1B}{2}$ (or just use the fact that $A_1B_1 = \frac{1}{2}AB)$.

As $AC_1 = C_1B, B_1M = M A_1$. So the intersection of line segments $CC_1$ and $A_1B_1$ is the midpoint of both segments.

Answer 2

Yes, they are equal.

Note that since $C_1$ is the midpoint of $AB,$ the vector $\vec{OC_1}$ must be half the sum of the vectors $\vec{OA}$ and $\vec{OB},$ in other words; $$\vec{OC_1}=\frac{\vec{CA}+\vec{CB}}{2}$$ $$\implies 2\vec{OC_1}=\vec{OA}+\vec{OB}$$ $$\implies 2\vec{OC_1}+2\vec{OC}=\vec{OA}+\vec{OB}+2\vec{OC}$$ $$\implies \frac{2\vec{OC_1}+2\vec{OC}}{4}=\frac{\vec{OA}+\vec{OB}+2\vec{OC}}{4}$$ $$\implies \frac{\vec{OC_1}+\vec{OC}}{2}=\frac{\vec{OA}+\vec{OB}+2\vec{OC}}{4}$$ $$\implies \frac{1}{2}\left(\vec{OC_1}+\vec{OC}\right)=\frac{1}{4}\left(\vec{OA}+\vec{OB}+2\vec{OC}\right).$$

Thus, they are equal, as required.