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See complete Problem 107Triangle, Angles. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
Construct E on AC such that angles ABC=BEC, then angles CBD=EBD and we have DE/BE=DC/CB=AB/BC so triangles ABC~DEB giving triangle BAD isosceles thus x=pi-4a.
Solution to problem 107.Take E on BC, such that AE is bisector of ang(BAC). Notice that ang(BAE) = ang(DBE) = alpha and both subtend the same segment DE. So ABED is cyclic, ang(BDE) = alpha, tr. BDE is isosceles and BE = DE.In triangle ABD, ang(BDC) = x + 2.alpha, soang(CDE) = x + alpha.As AB = CD, BE = DE andang(ABE) = ang(CDE) = x + alpha,triangles ABE and CDE are congruent, withang(BAE) = ang(DCE) = alpha.Thus, in triangle ABC,(x + alpha) + 2.alpha + alpha = 180and finally x = 180 – 4.alpha.
Let AE be the bisector of < BAC where E is on BCTr. s ABE and CDE are congruent SAS so < EDC = < ABE implying that ABED is cyclic and < DBC = @So x = 180-4@Sumith PeirisMoratuwa Sri. Lanka
Let E be on BC such that AE bisects < BACThen ABED is cyclic and so Tr.s ABE and DEC are congruent SASHence AE = EC and so < ACE = @So x = 180-4@Sumith PeirisMoratuwaSri Lanka
Bring AE=//DC (AB=DC) then AECD is parallelogram so <BEA=<BAE=<CAE=a then AE is bisector angle BAC.Is <ECB=<DBC=a=<BAE so BECA is isosceles trapezoid or concyclic.Therefore <BCA=<BAE=a and <BDA=2a,so x=180-4a.