## Sunday, June 14, 2009

### Problem 302. Triangle, Angle bisector of 60 degrees.

Proposed Problem
Click the figure below to see the complete problem 302 about Triangle, Angle bisector of 60 degrees.

Complete Problem 302
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

1. 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
1/b + 1/c = V3/x
Ajit Athle: ajitathle@gmail.com

2. Mrudul M. ThatteJune 16, 2009 at 9:45 PM

Let BC = a
So,
= 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,
= 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.

3. http://img392.imageshack.us/img392/7320/1001036z.jpg

4. draw DE//AB and DF perpendicular to AC.
win 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