We wanted to find ∠FEG in the crazy quadrilateral. Since ΔABC ~ ΔADE ~ ΔHAG, we know that ∠ABC = ∠ADE = ∠HAG = 90°. This means ∠BAC = 30°. We know that ΔAGE is isosceles. It is obviously impossible to have ∠EAG equal ∠AEG. It is also impossible for ∠EAG to equal ∠AGE, because then EF would intersect AH. Thus, ∠AGE = ∠AEG, so they must both measure 75°. ∠FEG = ∠AEF – ∠AEG, so we have our answer. ∠FEG measures 15°.