Proving by induction that an equilateral triangle will always be divided into (n+1)^2 small triangles? I'm working on a proof that looks like this:

Let $n$ be a positive integer. Given an equilateral triangle, place $n$ points on each side, dividing the side into $n+1$ equal segments.
Use the points to draw $n$ line segments parallel to each side of the triangle ($3n$ line segments in all).
Prove by induction that this will always divide the triangle into exactly $(n+1)^2$ little equilateral triangles.

But I am unsure of how to proceed past the base case.
 A: HINT: Your induction hypothesis is that when you divide the triangle with $n$ points on a side, and have $(n+1)^2$ small triangles. You want to show that when you divide it with $n+1$ points on a side, you get $(n+2)^2$ small triangles.
Imagine that you’ve divided it with $n+1$ points on a side. Show that if you ignore the bottom row of small triangles, the rest is an equilateral triangle that has been divided with $n$ points on a side. (You just have to check that its bottom edge has been divided with $n$ points into $n+1$ pieces.) By the induction hypothesis there are then $(n+1)^2$ small triangles above the bottom row of small triangles. Now show that the bottom row contains just enough small triangles to make up the difference between $(n+1)^2$ and $(n+2)^2$, i.e., $(n+2)^2-(n+1)^2$.

A: From a triangle with $(n+1)$-fold subdivision you can remove $n+1$ small triangles that have an edge on the base line and the $n$ triangles between them that have a vertex on the base line. The result is a triangle with $n$-fold subdivision.
Thus if $f(n)$ denotes the number of small triangles, we have
$$f(n+1)=f(n)+(n+1)+n.$$
A: Hint: 
If you split sides of $\triangle ABC$ in $n$ parts, then $\triangle AB'C'$ is a triangle whose edges were splitted in $(n-1)$ parts.
$\hspace{150pt}$
Cheers!
