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

\[PC :PB\]

.