MathLabs

Problem 1

Let P be a point in the interior of triangle ABC. Let D, E, F be the intersections of AP with BC, BP with CA, and CP with AB, respectively. Prove that the area of triangle ABC is 6 if each of triangles PFA, PDB, and PEC has area 1.
Step 2 of 5: Use the cevian through C
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yz=BFAF=[BPF][APF]=x−1,(z+1)x=x+y+z.\frac yz=\frac{BF}{AF}=\frac{[BPF]}{[APF]}=x-1,\quad (z+1)x=x+y+z.
Detailed analysis

Triangles BCP and ACP have bases on BC and AC and the same altitude from P, so their area ratio is BF:AF. The two triangles BPF and APF have the same altitude from P to AB, and [APF]=1 while [BPF]=x-1. Rearranging gives the displayed relation.