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Geometry ProblemClick the figure below to see the complete problem 516 about Triangle, Cevian, Congruence, Angles.
Let me give some supplement:Let point E, AE//DB and AD//EB.Triangle ACE is isosceles, angle EAC = angle BDC.So, angle AEC = angle ACE = 5x, that is, angle ECB = x.Angle EAB = x, since AE//DB.EA=BC, since points A, C, B, E are on a circle, and AC//EB.So, triangle ABE and CEB is congruence, that is,CB = EA = BD, so triangle DCB is isosceles ---> angle BDC is 6x.6x + 6x + 4x = 16x = 180deg. ---> x = 11.25deg.
Draw AE//BD with AE=BDADBE is a parallelogram and AEC isosceles . angle (EAC)=180-10x , angle(AEC)=5xAEBC is a cyclic quadrilateral with BE//AC so AEBC will become isosceles trapezoid with AE=BCAngle(BAC)=angle (ECA)=5xIn triangle ABC we have 5x+5x+6x=180 so x=11.25Peter Tran
Mark E on AC extended such that < CBE = x. Then DB = DE = AC so AD = CENow draw circle BDE to cut AB at P and BE at Q. Since equal arcs subtend equal angles in a circle PD = CQ which in turn subtend equal angles at Q and D showing that PQ//DCSo PDCQ is an isoceles trapezoid and Tr.s APD & CQE are congruent making Tr. ABE and hence Tr. BDC both isoceles. So < BDC = 6x and therefore 6x + 6x + 4x = 180 and x = 11.25Sumith PeirisMoratuwaSri Lanka
Problem 516Let E symmetry of D on the AB, then EB=BD=AC and <EBA=<DBA=x and <AEB=<ADB=10x.But<EBC=6x=<ACB then the AEBC is isosceles trapezoid (ΑΕ//ΒC).So <AEB+<EBC=180 or 10x+6x=180 therefore x=11.25APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE