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.