Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 663.
Friday, September 2, 2011
Problem 663. Scalene Triangle, 60 Degrees, Altitude, Angle Bisector, Perpendicular Bisector
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ABD is a right triangle with angle B=60
ReplyDeleteSo we get AF=FB=BD
Let 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 ABC
So AD, BF and FG concurrent at M
Peter Tran
Let BE and AD cut at M.
ReplyDelete∠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 = MB
Hence the ⊥ bisector of AB passes through M
If we draw MF perpendicular to AB, then Triangles AMF & BMF are congruent ASA
ReplyDeleteHence AF = FB so MF is the perpendicular bisector of AB
So the required concurrency is proved
Sumith Peiris
Moratuwa
Sri Lanka