# Coordinate free proof of height of intersection of two triangles with common base

Consider a triangle ABC where AB is given as the base and C is at height 1. (Note, I am fixing height as opposed to working with ratios because it is easier to state.) A point A' is placed on BC at height $a$, and a point B' is placed on AC at height $b$. The lines $AA'$ and $BB'$ intersect at a point $P$. What is the height of $P$?

By placing the triangle in the Cartesian plane and solving a linear system of equations, it is straight forward to show that the answer is $\frac{1}{\frac{1}{a}+\frac{1}{b}-1}$. However, it feels like there should be a way to compute this without resorting to fixing coordinates. Is there a way to solve this problem without resorting to coordinates?

To help make solutions more consistent, the line $CP$ intersects $AB$ at a point $C'$, and while this point isn't required to state the problem, I find it likely that it will make an appearance in any solution.

I will use the notation $[\cdot]$ for the area of a figure and also $d(P,\ell )$ for denoting the distance from point $P$ to line $\ell$. I will use this trivial fact: the areas of two triangles with equal heights are proportional to their respective bases.
By this fact, $$\frac{CA'}{A'B}=\frac{[AA'C]}{[AA'B]}=\frac{[PA'C]}{[PA'B]}=\frac{[AA'C]-[PA'C]}{[AA'B]-[PA'B]}=\frac{[PAC]}{[PAB]}.$$But $$\frac{[AA'C]}{[AA'B]}=\frac{[ABC]}{[AA'B]}-1=\frac{\frac12\cdot AB\cdot d(C,AB)}{\frac12 \cdot AB\cdot d(A',AB)}-1=\frac{1}{a}-1.$$So $$\frac{[PAC]}{[PAB]}=\frac1a-1.$$Similarly, $$\frac{[PBC]}{[PAB]}=\frac1b-1.$$ Adding them up, \begin{align*}\frac{[PAC]+[PBC]}{[PAB]}&=\frac1a+\frac1b-2\\ \implies\frac{[ABC]}{[PAB]}-1&=\frac1a+\frac1b-2\\ \implies \frac{\frac12\cdot AB\cdot d(C,AB)}{\frac12\cdot AB\cdot d(P,AB)}&=\frac1a+\frac1b-1\\ \implies\frac{1}{d(P,AB)}&=\frac1a+\frac1b-1.\end{align*}Therefore $d(P,AB)=\frac{1}{\frac1a+\frac1b-1}$ as desired. $\blacksquare$