Online Geometry theorems, problems, solutions, and related topics.
Proposed ProblemSee complete Problem 17 at:gogeometry.com/problem/problem017.htmLevel: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
From Tr. ABC, AB/AC = sin(2θ)AB = AC sin(2θ)=2ACsin(θ)cos(θ) = 2AC * DE/DC * DC/AC from Tr. CED & ADCThus, AB = 2DEAjit: email@example.com
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Nilton LapaTake F on de extension of BC, so that CF=AC. Tr. ACF is isosceles, with angle(CAF)=angle(CFA)=(theta). Let BC be the median of ACF. Draw GH perpendicular to CF. DAC and GCF are congruent so GH=DE. As F is middle point of CF, we have GH=AB/2 and AB=2.DE.
AB/DE = AB/AC. AC/CD . CD/DE = sin 2θ. sec θ. cosec θ = 2
Where I said "Let BC be the median of ACF", please change to "Let CG be ..."Where I said "As F is middle point of CF", please change to "As G is middle poit of AF".Sorry about that.Nilton Lapa
Solution is uploaded to the following linkhttps://docs.google.com/open?id=0B6XXCq92fLJJU2s3bDJBRDJBX2c
Take F - reflection of A in CD abd G projection of F onto AC. Easily FG=2DE and triangles CFG and CAB are congruent, hence FG=AB, done.Best regards,Stan Fulger
Let AC, DB meet at F and let AG drawn parallel to BC meet BD at G.ABCD is con cyclic and Tr.s ADF, BCF & AFG are all isoceles. Now Tr.s DEF & BAG being similar AB/DE = AG/EF = 2. AG/AF = 2 since AG=AFHence AB = 2.DESumith PeirisMoratuwaSri Lanka
Another idea: take F reflection of D in E; AFBCD is cyclic and we need DF=AB. Since CF is angle bisector of <ACB, there exists the equality BF=AF; but as constructed AF=AD, thus DAFB is an isosceles trapezoid, having equal diagonals, DF=AB, done.