# Ratio of area of a triangle to that of its medians

Let a triangle ABC have medians namely l,m,n. I can't figure out how to find the ratio of area of triangle LMN to area of triangle ABC. LMN is the triangle with sides of lengths of the three different medians of triangle ABC. I know that it is 0.75 but I cannot find its verification or proof, you name it. Well, I don't know where to start.

Would be great if you would lend me some help

PS: I am an Indian and may be my english is not that good, please ask to explain anything you do not understand....

• Is $\triangle LMN$ a triangle formed by three sides of lengths $l,m,n$? Jun 22, 2019 at 15:09
• l,m,n are the lengths of the medians Jun 22, 2019 at 15:09
• And triangle LMN is formed l,m,n respectively Jun 22, 2019 at 15:10

If the triangles $$T$$ and $$T'$$ are similar with simialrity factor $$k$$, then ratio of their areas is $$k^2$$. So since the sides of the starting triangle are twice as median triangle the result is $${1\over 4}$$.

Have I missed something?

Edit. You probably meant what is the area of a triangle formed by medians?

In that case, note that area of $$BDG$$ where $$D$$ is a midpoint of $$BC$$ is $${1\over 6}$$ of the $$ABC$$. Then if we reflect $$G$$ across $$D$$ we get twice as big triangle, which has sides $${2/3}$$ of the medians. So the ratio is $${1\over 3}\cdot ({3\over 2})^2 = {3\over 4}$$ • Sorry but I think I failed to explain the question. Jun 22, 2019 at 15:12
• However, I edited the question. I hope this makes it clear to you Jun 22, 2019 at 15:13
• How can I explain that BG' is equal to GC Jun 22, 2019 at 15:21
• If $C'$ halves $AC$ then $BG' = 2 CC' = GC$ Jun 22, 2019 at 15:22
• @Aqua You mean $C'$ halves $AB$. Jun 22, 2019 at 15:33