Proposed Problem
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
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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.
ReplyDeleteConsider ∆DAC:
ReplyDeleteAF is external bi-sector of ∠DAC (all angles 60°)
CF is internal bi-sector of ∠ACD (given)
Both these bi-sectors meet at point F
As external bi-sector of ADC is concurrent with above two bi-sectors (by theorem), DF is the external bi-sector of ∠ADC
=> ∠ADF = ½ of∠ADB -------- (1)
Similarly, consider ∆ADB, AE being external and BE being internal bi-sectors, DE must be external bi-sector of ∠ADB.
=> ∠ADE = ½ of∠ADC -------- (2)
Adding (1) and (2) we get
∠ADF+∠ADE = ½ of(∠ADB+∠ADC)
∠FDE = ½ of 180° = 90°
DE perpendicular to DF