## Tuesday, February 9, 2010

### Problem 424: Triangle, Cevian, Angle, 90 degree, Congruence

Proposed Problem
Click the figure below to see the complete problem 424 about Triangle, Cevian, Angle, 90 degree, Congruence. Complete Problem 424

Level: High School, SAT Prep, College geometry

1. http://geometri-problemleri.blogspot.com/2010/02/problem-70-ve-cozumu.html

2. Let be m(ACB)=y and m(BAC)=2y
in triangle ABC: 5y+x=90
knowing y is knowing x,and 0<5y<90
with the sine rule in ABD and ABC
BD=AB.sin2y/cos3y;BC=AB.sin2y/sin3y
in the right triangle BCD
BC²+BD²=DC²=AB²
then sin²(2y)=cos²(3y).sin²(3y)=sin²(6y)/4
we have the formula: sin(3a)=3sina-a(sina)^3
with t=2y ,sin²t=1/4
2y=30
x=15
.-.

3. Let E be the mid point of CD and F the mid point of AB. Then E is the centre of right Tr. BCD and so BF = BE.

Now < ABE = 180 - 2/3<C - 2<C = 180 - 8/3< C. Hence < BFE = < BEF = 4/3<C.

Hence < FEA = 4/3<C - 2/3<C = 2/3<C = < FAE. So Tr. BEF is equilateral and BE is perpendicular to AC and so < C = 45 and it follows that x = 15

Sumith Peiris
Moratuwa
Sri Lanka