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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 663.
ABD is a right triangle with angle B=60So we get AF=FB=BDLet AD cut FG at M .Triangle BFM congruence to triangle DDM ( case SAS)So angle FMB= angle DBM and BM is the angle bisector of angle ABCSo AD, BF and FG concurrent at MPeter Tran
Let BE and AD cut at M.∠ABM = 30°(since BE bisects ∠ABC = 60°)∠BAM = 30°(since it is the complement of ∠ABC=60° in the right-angled ∆ADB)So ∆AMB is isosceles and MA = MBHence the ⊥ bisector of AB passes through M