Geometry Problem

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

Click the figure below to see the complete problem 979.

## Wednesday, February 12, 2014

### Geometry Problem 979: Triangle, Cevian, Angle Bisector, Concurrency, 90 Degrees, Perpendicular

Labels:
90,
angle bisector,
cevian,
concurrent,
degree,
perpendicular,
triangle

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Applying Ceva’s theorem in triangle ABC

ReplyDeleteDA/DC x FC/FB x EB/EA= 1…….(1)

Replace DA/DC= DA/DB x DB/DC (2)

Since DE is angle bisector of ∠(ADB ) => DA/DB= EA/EB

Replace this in (2) and (1) and simplifying we have DB/DC x FC/FB= 1

Or DB/DC= FB/FC => DF is angle bisector of ∠(BDC)

∠(EDF)=1/2 ∠(ADC) = 90