Friday, September 2, 2011

Problem 663. Scalene Triangle, 60 Degrees, Altitude, Angle Bisector, Perpendicular Bisector

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 663.

 Online Geometry Problem 663. Scalene Triangle, 60 Degrees, Altitude, Angle Bisector, Perpendicular Bisector, Concurrent lines.

2 comments:

  1. ABD is a right triangle with angle B=60
    So 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

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  2. 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 = MB
    Hence the ⊥ bisector of AB passes through M

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