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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 view the complete problem 1042.
Let BC=k BE, then AB=k AD and AC=k AF. Let S(P, k, α) be the homothetic rotation transformation, with center P, scale factor k, and angle α. Consider first S(B, k, −α) and then S(A, 1/k, α). B→B→DE→C→FOn the other hand, the combined transformationS(A, 1/k, α)。S(B, k, −α) = T(v)which is a translation by vector v. [It is because (k)(1/k)=1 and (α)+(−α)=0]Hence, vector v = vector BD = vector EF. i.e. BEFD is a parallelogram.
Let DF cut BC at M and EF cut AB at N.Note that ABC is the image of ADF in the spiral similarity transformation withRotation angle = α => ∠(DMB)= α => DF//BEIn the similar way we also get EF//BD so DFEB is a parallelogram
Tr.s AFC and ADB are similar hence AF / b = AD / c = p say. So AF = pb and AD = pc and since < DAF = < BAC it follows that Tr.s ADF and ABC are similar the respective sides being in the ratio of pSo DF = pa = BE since tr.s AFC and BEC are similar. Similarly EF = BD = pc since Tr.s AFC and BEC are similar and Tr.s BEC and ABC are similarHence in quadrilateral BDFE the opposite sides are equal and hence BDFE is a parallelogram Sumith PeirisMoratuwaSri Lanka
Problem 1042The three equilateral triangles ADB ,BEC and AFC are similar to each other ,so AD/AB=BE/BC=AF/AC, But AD/AB=AF/AC or AD/AF=AB/AC with <DAF=α+<BAE=<BAC.Then the triangle DAF is similar with triangle BAC so AD/AB=DF/BC=BE/BC.ThereforeDF=BE. And triangle FCE is similar with triangle ACB so FE/AB=CE/BC=(BE/BC=AD/AB) soFE=DB.Therefore the DBEF is parallelogram.APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE