I've just started learning about similar triangles, and I'm having trouble with the following problem:
Let $T$ be the triangle $ABC$, as shown on the figure. Let $P$ be the midpoint of $\bar{AB}$ and $Q$ the midpoint of $\bar{AC}$. If $\bar{PQ}$ has length $5$ cm, find the length of $\bar{BC}$.
According to my book, we just have to note that triangle $ABC$ is the dilation by $2$ of triangle triangle $APQ$, and since $\bar{BC}$ is the side corresponding to $\bar{PQ}$, it follows that $\bar{BC}=10$ cm. However, I'm not satisfied with that answer at all. It hasn't been given to us that one triangle is a dilation of the other (the way I see it, that's precisely what we're trying to prove).
So I've drawn segment $\bar{AM}$ as on the figure below, and by use of Pythagoras' theorem and after some algebra I got the following equalities:
$$PQ=\sqrt{(AP)^2-(AN)^2}+\sqrt{(AQ)^2-(AN)^2}$$ $$BC=\sqrt{4(AP)^2-(AM)^2}+\sqrt{(4(AQ)^2-(AM)^2}$$
So basically I just need to show now that $\bar{AM}=2\bar{AN}$, but I'm not sure how to do that. Any help will be appreciated. Thanks.
by the intercept theorems we get $$PQ=\frac{1}{2}BC$$