Nobody has an answer for this, so I thought might as well post this on Jiskha. "Medians AX and BY of Triangle ABC are perpendicular at point G. Prove that AB=CG." I need an entire solution by Tuesday, May 20, 2014. I just need a little boost and thx for the help...

Answer 1

Nice problem involving
1. definition of sine
2. the cosine rule
3. the properties of medians, i.e. intersection of medians divides them in ratios of 2:1.

First need to do a diagram in words.
Given triangle ABC, with medians AX and BY meeting at G. AX and BY are orthogonal (i.e. perpendicular to each other).
Denote:
a=length of segments BX and XC
b=length of segments AY nd YC
2c=length of side AB.

Denote:
x=length of segment XG
y=length of segment YG

2x=length of segment GA
2y=length of segment GB
The last two statements stem from the properties of medians.

XY // BA (ΔCYX~ΔCAB, AA)
=> length of seg. YX = c
Use Pythagoras theorem for the three following statements.

Consider right triangle XGB:
x²+4y²=a² --(1)

Consider right triangle AGY:
4x²+y²=b² --(2)

Consider right triangle YGX:
x²+y²=c² --(3)

We will establish the cosines of angles CGY and CXG before applying the cosine rule to triangles CYG and CXG.

Lemma 1:
Cos(CXG)=-cos(BXG) (suppl. angles)
=-sin(XBG) (compl. angles)
=-x/a (definition of sine)

Lemma 2:
Similarly
Cos(CYG)=-y/b

Applying cosine rule to triangle CXG:
CG²=a²+x²-2xacos(CXG)
=a²+x²+2xasin(XBG)
=a²+x²+2x² (lemma 1)
=a²+3x² --(4)

Similarly,
CG²=b²+3y² --(5)

Add (4) and (5)
2CG²=a²+3(x²+y²)
=5(x²+y²)+3(x²+y*sup2;) (using (1) and (2))
=8(x²+y*sup2;)
=2(4c²)
=2(AB)²

Hence CG=AB.

Answer 2

Add (4) and (5)
2CG²=a²++b²+3(x²+y²)
=5(x²+y²)+3(x²+y*sup2;) (using (1) and (2))

Answer 3

Thank you MathMate! You're amazing!!!