Asked by da machine
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...
Answers
Answered by
MathMate
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.
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.
Answered by
MathMate
Add (4) and (5)
2CG²=a²+<b>+b²+</b>3(x²+y²)
=5(x²+y²)+3(x²+y*sup2;) (using (1) and (2))
2CG²=a²+<b>+b²+</b>3(x²+y²)
=5(x²+y²)+3(x²+y*sup2;) (using (1) and (2))
Answered by
Sheenybeany
Thank you MathMate! You're amazing!!!
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