ABCD is a parallelogram. E is on AD such that AE : ED = 1 : n - 1. BE meets the diagonal AC at P. If AP : AC = 1 : x, show that AB + nAE = xAP.
I proved it this way:
$$\frac{AP}{AC}=\frac{1}{x}$$ $$AC=xAP$$
and
$$\frac{AE}{ED}=\frac{1}{n-1}$$ $$ED=(n-1)AD$$
We know $$AD=AE+ED$$ $$AD=AE+(n-1)AE$$ $$AD=nAE$$
We can see that: $$AB=AC-DA$$ $$AB=xAP-nAE$$ Therefore, $$AB+nAE=xAP$$
It is the second part of the question that puzzles me a little:
Hence show that BE divides AC in the ratio of 1 : n
I am unable to see how I can use the previous result to show that.
$BE$ divides $AC$ in ratio $\frac{AP}{PC}$
since $\frac{AP}{AC}=\frac{1}{x}$, So $\frac{AP}{PC}=\frac{1}{x-1}$
Now we need to show that $x-1=n$
From your equation $$AB+nAE=xAP$$ Since three points $B,P$ and $E$ are collinear, sum of coefficients of $AB,AE,AP$ is equal to $0$ (see below for proof)
$\therefore 1+n-x=0$ so we showed that $x-1=n$
Proof:
Since $B,P,E$ are collinear
$$BP=\lambda PE$$
By triangle law $BP=AP-AB$ and $PE=AE-AP$, put it in above equation.
$$(AP-AB)=\lambda (AE-AP)$$ $$(\lambda)AE + (-1-\lambda)AP + (1)AB = 0$$
So you can notice that sum of coefficients is zero