Correct solutions were submitted by Bradley student Andrea Schoch and from proud BU alum Andrew Greene.  Correct solutions also came from Philippe Fondanaiche, France; Gary Smith, USA; Claudio Baiocchi, Italy; Bill Webb, USA;  Ron Welch, USA; Ali Riza Atasoy, Turkey; John Snyder, USA; Juan Carlos Marivela, Spain; Farid Lian, Colombia; Maria Luisa Cobo, Spain; Lou Cairoli, USA; Sam Collins, India.

The area of ACH = area of ACE - area of AHE = a(a + b)/2 - area of AHE.  So I'll find the latter area.  Drop a perpendicular from H to AE, meeting AE at the point P.  Let h = PH, x = PD.  Since CDE and PHE are similar triangles,

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Since AFE and AHP are similar triangles,

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Solve each equation for x, equate and solve to get 

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Hence 

area of ACH = (a + b)/2 -  (a + b)h/2 = .

As a worthwhile check note that this equation gives the area of ACH going to a²/2 as b goes to zero, which is geometrically accurate.

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