the question
Let the rectangle $ABCD$ with $AB=a\sqrt{2}, BC=2a$, E is the middle of the segment BC, and $F$ is the point of intersection of the lines AE and BD. Let M be the point such that $MA \perp (ABC)$ and $MA=a$. Find the measure of the angle formed by the planes $(MAE)$ and $(MBD)$.
The idea
The drawing
Because $BE || AD$ we get that $\frac{BE}{AD}=\frac{FE}{AF}=\frac{FB}{DF}=\frac{1}{2}$
Using the Pyathagorean Theorem, we get $$BD=a\sqrt{6},AE=a \sqrt{3} => AF=\frac{2a \sqrt{3}}{3}, DF=\frac{2a \sqrt{6}}{3}$$
Using the reciprocal of the Pythagoras theorem we get that $AF\perp BD$(1)
Using the Pythagorean Theorem we get $$MF=\frac{a \sqrt{21}}{3}$$ and using its reciprocal again we get that $MF \perp BD$(2)
From 1 and 2 and the fact that the intersection between the planes (MAE) and (MBD) we get that this planes are perpendicular, so the angle between them is 90.
I am not sure if my calculations are right because they do not seem perpendicular.