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$ABCD$ is a convex quadrilaterial such that $AC=BD$. $AC$ and $BD$ intersect at $E$ and $\angle AEB=66^{\circ}$. $F$ and $G$ are the midpoints of $AD$ and $BC$, respectively. $FG$ intersects $AC$ and $BD$ at $H$ and $I$, respectively. What is the measure (in degrees) of $\angle EHI$?

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Note: This is a problem posed on Brilliant.org in the previous week (when the user posted this problem). –  Calvin Lin May 14 '13 at 15:07

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The diagram

(three gray segments have same length)

Draw a parallelogram that can be made by CD and DB. Let B' the other point of the parallelogram. Then, ∠ACB' = ∠AEB = 66˚ and CB' = DB = AC.

Notice that G is also the midpoint of DB'. Therefore, FG and AB' are parallel.

Now, ∠EHI = ∠CAB'. Since CAB' is an isosceles triangle... ;)

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