Since the result does not depend on the shape of the triangle, we are free to choose this to make the maths easy. The following diagram shows the triangle that I used with the (x, y) coordinates of the triangle’s vertices and the points on each of its sides:
The equations for the two lines that cross at point P are:\[x=14-13y\] \[x=(4/3)y\] Solving these equations to find the y value at point P gives:\[y=42/43\] Since the triangles ABC and BCP have the same baseline, the ratio of their areas is equal to the ratio of their heights, which hence gives:\[\frac{\Delta BCP}{\Delta ABC}=(42/43)/7 = 6/43\] The triangle ABC is made up of the central triangle and three triangles each with \(6/43\) of its area, which means that the central triangle has an area of \(25/43\) of its area. And knowing that the area of the central triangle is \(100 cm^2\) allows us to derive the area of triangle ABC as \(172 cm^2\).

Routh’s theorem deals with the more general case in which the sides are divided by the cevians in different proportions. Here is a short note deriving the area relationships for the general case.