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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to enlarge the figure of problem 1014.
Since ∠ (EAH)= ∠ (EBH)=45 => ABEH is cocyclicAnd ∠ (EHA)= 90 => AEH is a right isosceles triangleAE/AH=sqrt(2)Similarly we also have AF/AG= sqrt(2)Triangle AGH similar to triangle(AFE) with ratio of similarity= sqrt(2)And area(AEF)=2. Area(AGH)So S1=S2
It suffice to prove that (AE×AF)/(AG×AH) = 2. Using trigonometry, Let ∠BAE=α and ∠DAF=β, then α+β=45°. (AE×AF)/(AG×AH)= AE/AG × AF/AH= (AB+BE)/AB × (AD+DF)/AD --- Angle bisector theorem= (1 + BE/AB) × (1 + DF/AD)= (1 + tanα) × (1 + tanβ)= 1 + tanα tanβ + tanα + tanβ= 1 + tanα tanβ + tan(α+β) [1 − tanα tanβ] --- Compound angle formula= 1 + tanα tanβ + 1 − tanα tanβ= 2
AGFD is cyclic GF is perpendicular to AG. Let AG = GF = a and let GE = bABEH is cyclic so AH is perpendicular to EH and AH = EH = (a+b)/sqrt2S1 = S(AGF) X AH/AF = a^2/2 X (a+b)/sqrt2 / (sqrt2 a)= a(a+b)/4...(1)S1+S2 = 1/2 a(a+b)...(2)From (1) and (2) S1 + S2 = 2S1 and hence S1 = S2Sumith PeirisMoratuwaSri Lanka