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Click the figure below to see the complete problem 368 about Triangle, 120 degrees, Angle bisectors, Perpendicular.

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Complete Geometry Problem 368

Level: High School, SAT Prep, College geometry

## Saturday, October 17, 2009

### Problem 368. Triangle, 120 degrees, Angle bisectors, Perpendicular

Labels:
120,
60,
angle bisector,
perpendicular,
triangle

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In Tr.ABC, AC being the angle bisector and A=120 deg., we've AD=2bc*cos(A/2)/(b+c)=bc/(b+c).

ReplyDeleteNow let A be (0,0). This makes D:[bc/2(b+c),V3bc/2(b+c)] and E:[bc/(a+b),0] while F becomes [-bc/2(b+c),V3bc/2(c+a)] where V3 = square root of 3.

Slope DE=(V3bc/2(b+c))/[bc/2(b+c)-bc/(b+c)]=V3(a+b)/(a-b-2c) on simplification. Similarly it can be shown that slope DF=V3(a-b)/(a+b+2c).

Finally, slope DE*slope DF=3(a^2-b^2)/[a^2-(b+2c)^2). However, a^2=b^2+c^2+bc by co-sine rule in Tr. ABC. Hence the product of the slopes=3(c^2+bc)/(c^2+bc-4bc-4c^2)= -1 which means that DE and DF are pependicular to each other.

Ajit: ajitathle@gmail.com

In ∆ ABD,

ReplyDeleteBE interior bisector of angle B,

AC exterior bisector of angle A,

BE intersects AC at E

therefore

DE bisects angle ADC

In ∆ ACD,

CF interior bisector of angle C,

AB exterior bisector of angle A,

CF intersects AB at F,

Therefore

DF bisects angle ADB

DE perpendicular to DF

Essafty on what you base your proof on? coz its unclear to me how is DE and DF bisectors.

ReplyDeleteEssafty uses the fact that the 3 angle bisectors are concurrent.

ReplyDelete