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Proving geometric identities for special hexagon

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October 10, 2010
trigonometrygeometrycircumcircletrig identitiesLaw of Sinesgeometry unsolved

Problem Statement

The opposite sides of the reentrant hexagon AFBDCEAFBDCE intersect at the points K,L,MK,L,M (as shown in the figure). It is given that AL=AM=a,BM=BK=bAL = AM = a, BM = BK = b, CK=CL=c,LD=DM=d,ME=EK=e,FK=FL=fCK = CL = c, LD = DM = d, ME = EK = e, FK = FL = f. http://imgur.com/LUFUh.png (a)(a) Given length aa and the three angles α,β\alpha, \beta and γ\gamma at the vertices A,B,A, B, and C,C, respectively, satisfying the condition α+β+γ<180∘\alpha+\beta+\gamma<180^{\circ}, show that all the angles and sides of the hexagon are thereby uniquely determined. (b)(b) Prove that 1a+1c=1b+1d\frac{1}{a}+\frac{1}{c}=\frac{1}{b}+\frac{1}{d} Easier version of (b)(b). Prove that (a+f)(b+d)(c+e)=(a+e)(b+f)(c+d)(a + f)(b + d)(c + e)= (a + e)(b + f)(c + d)
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