Let ABCD be a trapezoid; find point M on AB, such that $DM+MC$ is minimal possible Let ABCD be a trapezoid with $A=B=90$, $AD=a$ and $AB=BC=2a$.
Find point M on AB, such that $DM+MC$ is minimal possible.

I have been trying to do this question, but without success. I have succeeded in proving that $DA+AC<DB+BC$ something which I believe would help in finding this minimal value, however I have not succeeded in finishing it off. Moreover, I have also found that from Pythagoras we have that $DA+AC=a+2a\sqrt{2}$ and $DB+BC=2a+a\sqrt{5}$. Can someone please help me finish the question off?

Another way to solve -
Choose a point $M$ on $AB$ such that $AM = x, \,$ then BM = $(2a-x)$
$DM = \sqrt{AD^2 + AM^2} = \sqrt{a^2+x^2}$
Similarly $MC = \sqrt{BC^2 + BM^2} = \sqrt{4a^2+(2a-x)^2}$
$L = DM + MC = \sqrt{a^2+x^2} + \sqrt{8a^2+ x^2 - 4ax}$
To find min value of $L$,
$ \displaystyle \frac{dL}{dx} = \frac {x} {\sqrt{a^2+x^2}} + \frac{x-2a}{\sqrt{8a^2+ x^2 - 4ax}} = 0$
$\displaystyle x \sqrt{8a^2+ x^2 - 4ax} + (x-2a) \sqrt{a^2+x^2}  = 0$
$\displaystyle x \sqrt{8a^2+ x^2 - 4ax} =  (2a-x) \sqrt{a^2+x^2}$
$\displaystyle x^2 (8a^2+ x^2 - 4ax) =  (2a-x)^2 (a^2+x^2)$
Solving we get,
$ \displaystyle (x + \frac{2a}{3})^2 = \frac{16a^2}{9}$
So, $x = \frac{2a}{3}$ or $x = -2a$
Discarding $x = -2a$, $AM = x = \frac{2a}{3}$ gives you min.
EDIT: Just added a diagram for what Calvin Lin suggested which uses reflection -
$DM + MC$ will minimize  when it is a straight line.

