Solving the following relation in triangle

Question

If a line through the centroid $G$ of a triangle ABC meets $AB$ in $M$ and $AC$ on $N$ then prove that $AN. MB+AM. NC=AM. AN$ both in magnitude as well as sign

I tired dividing the equation by $AM. AN$ Thus resultant became $\frac{MB}{AM}+\frac{NC}{AN}=1$ and now I tried to do some construction but couldn't figure out the solution.

score 1 · Answer 1 · answered Jul 30 '14 at 11:49

1

Hint: Draw perpendicular lines to line $MN$ through $A$, $B$, $C$, and the midpoint $D$ of $BC$. Turn the ratios you got into ratios of the length of the perpendicular segments.

answered Jul 30 '14 at 11:49

Quang Hoang

15,854

Could you please elaborate – Apoorv Jain Jul 30 '14 at 12:33
You want to prove that the distance from $A$ to line $MN$ is equal to the total distances from $B$ and $C$ to line $MN$. Do you see it? – Quang Hoang Jul 30 '14 at 12:34
I got this i used the idea that you gave of drawing perpendiculars and then I used similarities but how can I further prove that the distance from A to MN is equal to the sum of the distances from B to MN and from C to MN please help. Thanks – Apoorv Jain Jul 30 '14 at 12:46
How about the perpendicular from $G$? How does it relate to the other perpendiculars? – Quang Hoang Jul 30 '14 at 12:48
I ain't getting it please present give a solution – Apoorv Jain Jul 30 '14 at 12:49
Using similarity, you should be able to show the distance from $A$ is twice the distance from $G$. – Quang Hoang Jul 30 '14 at 12:51
Yeh I got that and then – Apoorv Jain Jul 30 '14 at 12:53
Don't know how to do so – Apoorv Jain Jul 30 '14 at 12:57
Now use the fact that $D$ is the midpoint of $BC$ to show that the distance from $D$ is half of the total distance from $B$ and $C$. – Quang Hoang Jul 30 '14 at 12:58
Please explain and give a full solution please please – Apoorv Jain Jul 30 '14 at 13:00

score 0 · Answer 2 · answered Mar 08 '22 at 00:23

Let $X$ be the intersection of the lines $MN$ and $BC$. Here is a picture:

Apply the theorem of Menelaus for the triangles $\Delta ABD$, $\Delta ADC$ to get (using sign conventions for proportions of two segments on the same line):

$$ \begin{aligned} 1 &= \frac{MB}{MA}\cdot \underbrace{\frac{GA}{GD}}_{=-2}\cdot \frac{XD}{XB}\ , \\ 1 &= \frac{NC}{NA}\cdot \underbrace{\frac{GA}{GD}}_{=-2}\cdot \frac{XD}{XC}\ , \qquad\text{ and from here} \\[3mm] \frac{MB}{MA} + \frac{NC}{NA} &=-\frac 12\left( \frac{XB}{XD} + \frac{XC}{XD}\right) = -\frac 12\cdot\frac{XB+XC}{XD}= -\frac 12\cdot\frac{2XD}{XD} =\boxed{\ -1\ }\ . \end{aligned} $$ $\square$

Note: By sign convention we have $\displaystyle \frac{MB}{MA} = -\frac{MB}{AM}$, and $\displaystyle \frac{NC}{NA} = -\frac{NB}{AN}$, making the sign a plus one for the sum in the question.

Solving the following relation in triangle

2 Answers2