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...
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.
Add (4) and (5)
2CG²=a²++b²+3(x²+y²)
=5(x²+y²)+3(x²+y*sup2;) (using (1) and (2))
Thank you MathMate! You're amazing!!!
To prove that AB=CG, we need to use the fact that medians of a triangle divide each other in a 2:1 ratio.
Here's a step-by-step solution:
1. Draw triangle ABC with medians AX and BY intersecting at point G.
2. Let M be the midpoint of side BC, and N be the midpoint of side AC.
3. Since AX is a median, it passes through the midpoint M of side BC. Thus, AM is half the length of BC.
4. Similarly, BY is a median, so it passes through the midpoint N of side AC. Hence, BN is half the length of AC.
5. Since M is the midpoint of BC, we have MB = MC. Similarly, since N is the midpoint of AC, we have AN = NC.
6. Since AX and BY intersect at G, it implies that G divides AX into AG and GX in a 2:1 ratio, and it divides BY into BG and GY in a 2:1 ratio as well.
7. Since AG and BG are two-thirds of the lengths of the medians AX and BY, respectively, we can express them as AG = (2/3) * AM and BG = (2/3) * BN.
8. Since AB is equal to the sum of AG and BG, we have AB = AG + BG. Substituting the values of AG and BG, we get AB = (2/3) * AM + (2/3) * BN.
9. But AM = MB/2 and BN = NC/2, so AB = (2/3) * (BC/2) + (2/3) * (AC/2).
10. Simplifying further, AB = BC/3 + AC/3, which can be rewritten as AB = (BC + AC)/3.
11. Now, since G is the intersection of medians AX and BY, it divides both medians into segments in a 2:1 ratio. This means that CG is two-thirds of AX and two-thirds of BY.
12. Therefore, we can write CG as CG = (2/3) * AX and CG = (2/3) * BY.
13. Since AX and BY are both medians, they have the same length as their corresponding sides. Thus, AX = BC and BY = AC.
14. Substituting these values, we get CG = (2/3) * BC and CG = (2/3) * AC.
15. Adding BC/3 and AC/3, we have (2/3) * BC + (2/3) * AC, which simplifies to (2/3) * (BC + AC).
16. But (BC + AC)/3 is equal to AB, as we proved in step 10. Therefore, CG = AB.
Hence, we have proven that AB = CG.