Proposed Problem
Click the figure below to see the complete problem 302 about Triangle, Angle bisector of 60 degrees.
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Complete Problem 302
Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Sunday, June 14, 2009
Problem 302. Triangle, Angle bisector of 60 degrees.
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Tr. ADC and Tr. ADB give us: x/b = sin(C)/sin(B+30) & x/c = sin(B)/sin(C+30). Thus, x/b+x/c= sin(C)/sin(B+30) + sin(B)/sin(C+30) But angles B+30 and C+30 are supplementary. Thus,x/b+x/c= (sin(C)+ sin(B))/sin(B+30). Further, sin(C)=sin(B+60)=sin(B)/2+V3cos(B)/2. So, x/b + x/c = [sin(B)+sin(B)/2+V3cos(B)/2]sin(B+30) which on simplification = V3. Hence, x/b + x/c = V3 or
ReplyDelete1/b + 1/c = V3/x
Ajit Athle: ajitathle@gmail.com
Let BC = a
ReplyDeleteSo,
AD*AD = b*c*{ 1 - [a*a / (b+c)*(b+c)]}
= b*c*{ b*b + 2*b*c + c*c - a*a }/
[(b+c)*(b+c)]
As angle BAC = 60,
a*a = b*b + c*c - b*c
So,
AD*AD
= b*c* {b*b + 2*b*c + c*c - b*b - c*c + b*c}/ [(b+c)*(b+c)]
= b*c*{3*b*c} /[(b+c)*(b+c)]
So,
AD = [Square root of 3]*b*c /[b+c]
So,
x/[Square root of 3] = b*c/[b+c]
So,
[Square root of 3]/x = [b+c] / [b*c]
= [1/b] + [1/c]
Hence the proof.
http://img392.imageshack.us/img392/7320/1001036z.jpg
ReplyDeletedraw DE//AB and DF perpendicular to AC.
ReplyDeletewin DE=AE, DF=x/2
tr DEC and tr ABC are similar (AB//DE), =>DE/c = b-DE/b
=> DE = bc/b+c
from right tr DEF => DE'2 - DE'2/4=x'2/4 => DE=x/v3
sub. x/v3=bc/b+c give the result
p.s. '2 mean exponent 2
v3 mean cube root
Area(ABC)=Area(ABD)+Area(ADC)
ReplyDelete½*b.c.sin(60)=1/2*c.x.sin(30)+1/2*b.x.sin(30)
So b.c.SQRT(3)=b.x+c.x
Divide both side by b.c.x we will get the result
Peter Tran
Altitude BH = sqrt3/2c and altitude DG = x/2. Now use the fact that Tr.s BHC & DGC are similar and that HC = b-c/2 and GC = b- xsqrt3 /2 and the result follows
ReplyDeleteSumith Peiris
Moratuwa
Sri Lanka