See complete Problem 174 at:
www.gogeometry.com/problem/p174_quadrilateral_area_midpoint.htm
Quadrilateral with Midpoints, Triangles, Areas. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Sunday, September 7, 2008
Elearn Geometry Problem 174
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[BAF]+[EDC]=[BFD]+[BDE]=[BFE]+[EFD]=[EFC]+[EAF]=[EAFC]
ReplyDeletethen S=S_1+S_2 is obvious
Let AD=2a, and the distances from B, E and C to AD be b, e and c respectively.
ReplyDeleteWell known, e=(b+c)/2 ( 1 ) (trapezoid midline property). Multiply each side of (1) by a, to get ae=(ab+ac)/2, i.e. [AED]=[AFB]+[FCD], where from the required equality is obvious.
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Define G the intersection of AE and BF and H the intersection of DE and CF
ReplyDeleteDefine M=[BGE], N=[EHC], P=[AGF], Q=[FHD], Sg=[GEF], Sh=[HEF] with Sg+Sh=S
(1) : [ABCD]=[ABD]+[CBD]=2[ABF]+2[CDE]=2(S1+P+S2+N)
P+Sg=Q+Sh and M+Sg=N+Sg => P-M=Q-N therefore (3) : P+N=Q+M
(2) : [ABCD]=S1+S+S2+M+N+P+Q
(1) and (2) : S1+S+S2+M+N+P+Q = 2(S1+P+S2+N)
=> S+M+Q= S1+P+S2+N
But (3) : P+N=Q+M => S=S+S2