Proof That the Area of a Triangle Cut at a Vertex is Equal to the Ratio in Which the Opposite Side is Cut

Given an arbitrary triangle, we can draw a line from a vertex to the opposite side, and the areas of the resulting triangles is in the ratio in which that side is cut.

Draw a line from the top vertex to cut the base at P.

The are of APC is
$\frac{1}{2} PC \times PA \times sin (\angle APC)$
.
The are of APB is
$\frac{1}{2} PB \times PA \times sin (\angle AP)$
.
The ratio of the areas is
$\frac{1}{2} PC \times PA \times sin (\angle APC): \frac{1}{2} PB \times PA \times sin (\angle AP)$
.
$PC \times sin (APC):PB \times sin (\angle AP)$
.
But
$sin (\angle APC)=sin (\angle AP)$
since
$\angle APC+ \angle APB=180$
.
Hence the ratio of the areas of the triangles is
$PC :PB$
.