ABCD is a square, E a point onto AD, F onto BC so that BE||DF. CE x AB=G, GF x AD = H. Prove that <BEC=<FHE.
Proof:
Let AB=a. We have AE=CF ( 1 ), DE=BF ( 2 ). Take K = CH x BE, L = AF x BE.
From AD||BC: HE/AE=CF/BC; with (1), HE/AE=AE/a, giving a.HE=AE^2 ( 3 ). Divide both sides of (3) by a^2: HE/a=AE^2/a^2=AE^2/AB^2 ( 4 ), but HE/a=HE/BC=KE/BK ( 5 ), implying
KE/BK=(AE/AB)^2, thus AK_|_BE ( 6 ).
From similar triangles AED and ALE: AE/AD=AL/AF ( 7 ), while from triangles ALE and FLB similar we get AL/AF=EL/BE ( 8 ). But again HE/AH=CF/BF=AE/BF=EL/BL, hence LH||AB ( 9 ). From (6) and (9) ALKH is cyclic, so <ALE=<KHE, but <ALE=<CEB, done.
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