Given $\triangle ABC$ isosceles with base AB, M and N the midpoints of the equal sides. Show the reflection of C is the midpoint of AB

Given $\triangle ABC$ isosceles with base AB, M and N the midpoints of the equal sides. Show the reflection of C is the midpoint of AB. enter image description here

These are some observations I have made (although I can't use them in my favor):

We can write $\triangle C'MN$, and It will be congruent to $\triangle CMN$ (and form 4 congruent triangles);
$C' \in$ the height of $\triangle ABC$;

Solution 1:

I'll take for a fact that $MN \parallel AB$. You may choose to see it as a consequence of triangle similarity or Thales' intercept theorem.

Let $P = CC' \cap MN$.

We know from the definition of reflection with respect to lines that $\angle MPC = 90^{\circ}$ and that $PC = PC' = \frac{CC'}2$. We also know that $\triangle MCP \sim \triangle ACC'$ because of SAS in $\angle MCP$, so:

$AC' \parallel MP \parallel MN \parallel AB \implies C' \in AB$

from the same similarity we get that $\angle AC'C = 90^{\circ}$ so $C'$ is the foot of the height from $C$ to $AB$, which can be easily shown to be the midpoint of $AB$ since $\triangle ACB$ is $C$-isoscelis: $\triangle ACC' \cong \triangle BCC'$ by SAS: they share $\angle ACC' = \angle BCC', CB= AC$ and $CC'$ is common, so $C'A = C'B$.