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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1194.
Draw MN middle line of ADEC (= 12)Apply theorem of pythagore MNB, ADB, BECwe get x=14
Draw CF ⊥ DA ..( F on DA)Let z= x^2Applying Pythagoras’s theorem in triangles ABD, BCE and ACF we get BE^2=x^2-121=z-121BD^2=z-169AC^2=CF^2+AF^2 => (BE+BD)^2+4=z Or 2.BE.BD=z-BE^2-BD^2-4= 286-z4.BE^2.BD^2=4(z-121)(z-169)=(286-z)^2Simplify this equation we get 3.z^2-588.z= 0We get 2 solutions z= 0 and z= 196Only z=196 is acceptable so x= 14
See below for geometry solution of problem 1194http://s30.postimg.org/urgsbgadt/pro_1194.pngDraw circles diameters BA and BCDraw altitudes BG of tri. ABC and GH of triangle DGEObserve that H is the midpoint of DE => DGE is isosceles tri.∡ (DGE)= ∡ (DGB)+ ∡ (BGE)= ∡ (DAB)+ ∡ (BCE)= 180-120= 60 degreesSo DGE is an equilateral tri.HG= ½( DA+CE)= 12= DE. √ (3)/2So DE= 8. √3=CFIn right triangle ACF we have x^2= AF^2+CF^2= 4+(8√3)^2=196 => x= 14
DB + BE = √(x²-（13-11）²)√(x²-13²) + √(x²-11²) = √(x²-（13-11）²)x = 14
Sumith Peiris has left a new comment on your post "Geometry Problem 1194: Equilateral Triangle, Perpe...":Applying Pythagoras 3 times we seeDE = sqrt(x^2-13^2) + sqrt(x^2-11^2)=sqrt(x^2-2^2)One can solve this equation by algebra but the equation becomes somewhat unwieldyBy trial and error x = 14 (check: sqrt27 + sqrt75 = sqrt(27 + 75 + 2Xsqrt(27X75)= sqrt(102 + 2 X 45)=sqrt192
Let F be the mid point of AC. Easily ADBF and BECF are concyclic and DFE is equilateral Use Ptolemy in these 2 concyclic quadrilaterals and if DB = a and BE = b derive that a+b = 8sqrt3. So x^2 = (13-11)^2 + (8sqrt3)^2 = 196 and so x = 14Sumith PeirisMoratuwaSri Lanka
MN middle line of ADEC, CP perpend AD. ∆MBN ~ ∆CPN => BN = √3Pyth Th for BNM => 3+144=(3x²)/4 => x=14