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

Labels:
60 degrees,
altitude,
angle,
angle bisector,
concurrent,
perpendicular bisector,
scalene,
triangle

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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