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  
\[\frac{1}{2} PC \times PA \times sin (\angle APC): \frac{1}{2} PB \times PA \times sin (\angle AP)\]
\[PC \times sin (APC):PB \times sin (\angle AP)\]
\[ sin (\angle APC)=sin (\angle AP)\]
\[\angle APC+ \angle APB=180\]
Hence the ratio of the areas of the triangles is  
\[PC :PB\]

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