Using Ceva's Theorem Proof on Area of a Triangle

Question

I am having trouble identifying the height of each triangle.

Answer 1

Hint: triangles with bases on the same line and a common third vertex have areas proportional to their bases (because they share the same height). Use that to prove that:

$$\frac{AC'}{C'B} = \frac{area(\triangle CAC')}{area(\triangle CC'B)} = \frac{area(\triangle PAC')}{area(\triangle PC'B)}$$

Then note that $area(\triangle CPA) = area(\triangle CAC') - area(\triangle PAC')$, and similar for $\triangle CPB$.

Finally, remember that for equal fractions (with $b \ne d)$:

$$\lambda = \frac{a}{b} = \frac{c}{d} \quad \implies \quad \lambda = \frac{a-c}{b-d}$$

So:

$$\frac{AC'}{C'B} = \frac{area(\triangle CAC') - area(\triangle PAC')} {area(\triangle CC'B) - area(\triangle PC'B)} = \frac{area(\triangle CPA)}{area(\triangle CPB)}$$

which concludes the proof.