Points $D$, $E$, and $F$ are the midpoints of sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively, of $\triangle ABC$. Points $X$, $Y$, and $Z$ are the midpoints of $\overline{EF}$, $\overline{FD}$, and $\overline{DE}$, respectively. If the area of $\triangle XYZ$ is 21, then what is the area of $\triangle CXY$?
See the following image :
Note that XY is parallel to ED, YZ is parallel to DC and XZ is parallel to EC .... therefore triangle XYZ is similar to triangle EDC
And since X,Y bisect FE and FD in triangle EFD and XY is parallel to ED then
Triangle FXY is similar to triangle FDE...and....
XY / FY = ED / 2FY
XY = ED / 2
2XY = ED
And since XYZ is similar to EDC.....the altitude of EDC = 2*altitude of XYZ
And since XY, ED and IH are parallel.....the altitude of triangle EDC will be the same altitude as in triangle IHC
So the altitude of triangle CXY =
altitude of triangle XYZ + altitude of triangle EDC =
altitude of triangle XYZ + 2* altitude of triangle XYZ =
3*altitude of triangle XYZ
And triangle CXY is on the same base as triangle XYZ....so their areas are to each other as their altitudes
So [ CYX ] = 3 * [ XYZ ] = 3(21) = 63