Equilateral $\triangle ABC$ has side length $2$, $M$ is the midpoint of $\overline{AC}$, and $C$ is the midpoint of $\overline{BD}$. What is the area of $\triangle CDM$? $\textrm{(A)}\ \frac {\sqrt {2}}{2}\qquad$ $ \textrm{(B)}\ \frac {3}{4}\qquad$ $ \textrm{(C)}\ \frac {\sqrt {3}}{2}\qquad$ $ \textrm{(D)}\ 1\qquad$ $ \textrm{(E)}\ \sqrt {2}$
Solution
Connect $BM$. Triangles $CDM$ and $CBM$ have equal bases $CD=CB$ on the same line and the same height from $M$, so $[CDM]=[CBM]$.
$M$ is the midpoint of $AC$, so $[CBM]=\frac12[ABC]=\frac12\cdot\frac{\sqrt3}4\cdot 2^2=\frac{\sqrt3}2$.
Equilateral $\triangle ABC$ has side length $4$, and $M$ is the midpoint of $\overline{AC}$. Points $B,C,D$ are collinear, with $C$ between $B$ and $D$ and $CD=2BC$. What is the area of $\triangle CDM$?
$\textrm{(A)}\ \frac {\sqrt {2}}{2}\qquad$ $ \textrm{(B)}\ \frac {3}{4}\qquad$ $ \textrm{(C)}\ \frac {\sqrt {3}}{2}\qquad$ $ \textrm{(D)}\ 1\qquad$ $ \textrm{(E)}\ \sqrt {2}$ 
