# Prove that in similar triangles ratio of correspondent medians is same as ratio of correspondent sides

I had a math exam today about geometry and similar triangles. One of our math puzzle wanted us to proves something. Now I’ll explain that for you and if you help me I won’t lose 2 points of my midterm exam! So imagine that I’m your student and you’ve asked this question and I’ve answered like that.

QUESTION: We have two similar triangles. Prove that ratio of correspondent medians is same as ratio of correspondent sides.

MY ANSWER: We suppose two similar triangles, $$\triangle ABC$$ and $$\triangle A’B’C’$$. and also I did not mention that sides are equal! I mean $$AB \neq A’B’$$ , $$AC \neq A’C’$$ , $$BC \neq B’C’$$. And then I draw diagram 2. You can take a look here.

Actually I combined shapes in diagram 1, and I just draw diagram 2 in my exam paper. (I changed name of points in diagrams to explain what I answered better)

I wrote that we know: $$\triangle ABC\thicksim \triangle AMN$$ $$MN \parallel BC$$ $$BH=HC$$ $$MO=ON$$ $$AO \space, AH$$ are medians

So I continued based on thales theorem: $$\frac{AM}{MB}=\frac{AO}{OH}$$ $$\frac{AN}{NC}=\frac{AO}{OH}$$ Thus $$\frac{AM}{MB}=\frac{AN}{NC}$$ On the other side : $$\frac{AM}{MB}=\frac{AN}{NC}=\frac{AO}{OH}$$ And finally he gave me a big beautiful zero! I don’t know why and I hadn’t a change to talk to him. What’s your Idea? Is my answer OK? If yes tell me why. Because I’m going to convince him.

Staying with diagram 1, by similarity$$\frac{AB}{BC}=\frac{A'B'}{B'C'}$$But since $$M$$ and $$M'$$ are midpoints we know$$\frac{AB}{BM}=\frac{A'B'}{B'M'}$$And we're given $$\angle B=\angle B'$$Therefore (Euclid, Elements VI, 4 & 6)$$\triangle ABM\sim\triangle A'B'M'$$from which it follows that$$\frac{AB}{A'B'}=\frac{AM}{A'M'}$$

I'm not sure it helped to superimpose the triangles (diagram 2). Your concluding proportion has lines $$MB$$, $$NC$$, $$OH$$, which are differences of corresponding sides, but doesn't the question call for $$AB$$, $$AC$$, $$AH$$ instead? You're almost there, and the teacher's zero seems a bit harsh, but you may not win your case.

• Yep! I agree with you Mr.Porcella. I actually hadn't any Ideas about AB , AC , and AH and still no! Because our teacher didn't prove that and when i talked to one of my classmates his idea was that. – user602338 Nov 22 '18 at 7:10
• Mr Porcella you are a teacher. Actually you are a retired one! So as a student who is going to study a lot in chemistry, physics, biology, literature , english, arabic, geography and so on for the entrance exam to medical college, is it reasonable that I spare my time with these math proofs that does'nt really help me!? – user602338 Nov 22 '18 at 7:14
• Well you've got a lot to study, but mathematical proofs can help train you in ways that might not be obvious at first. Someone asked Abraham Lincoln once how he learned to think and write so clearly, and he said it was by studying the proofs in the first six books of Euclid's Elements. All the best to you. – Edward Porcella Nov 22 '18 at 16:18
• Yeah surely I agree with you there! Good night! – user602338 Nov 22 '18 at 18:07

I think you have a reasonable idea here, but your proof is incomplete. In order to apply Thales' theorem in this way, you need to know that $$A$$, $$O$$, and $$H$$ are collinear, and you haven't given any reason why they should be.

Notice that you haven't ever used the fact that $$O$$ and $$H$$ are midpoints. This is what you will need to prove collinearity.

• SO WOULD YOU TELL ME HOW CAN WE PROVE TGAT THEY ARE COLLINEAR? – user602338 Nov 19 '18 at 17:26
• APPLY THALES' THEOREM AGAIN TO SHOW THAT THE INTERSECTION OF $\overline{AH}$ WITH $\overline{MN}$ IS THE MIDPOINT OF $\overline{MN}$, AND THUS MUST COINCIDE WITH $O$. (also, stop shouting) – Micah Nov 19 '18 at 17:29
• Oh sorry i did't shout! My keyboard capslock was on! Also im not a native formal speaker bro hh! :D – user602338 Nov 19 '18 at 17:39