Area of Triangle in Terms of Side and Altitude

Theorem
The area of a triangle $$ABC$$ is $$\frac{c\cdot h_c}{2}=\frac{b\cdot h_b}{2}=\frac{a\cdot h_a}{2}$$ where $$a$$,$$b$$, and $$c$$ are sides and $$h_a$$, $$h_b$$ and $$h_c$$ are the altitudes from $$A$$,$$B$$ and $$C$$ respectively.

=Proof=

Construct a point $$D$$ so that $$\Box ABDC$$ is a parallelogram.

Then we have $$\triangle ABC \cong \triangle DCB$$, hence their areas are equal.

The area of a parallelogram is equal to the product of one of its bases and the associated altitude.

Thus $$(ABCD) = c\cdot h_c \Longrightarrow 2(ABC)=c\cdot h_c \Longrightarrow(ABC)=\frac{c\cdot h_c}{2}$$ where $$(XYZ)$$ is the area of the plane figure $$XYZ$$.

A similar argument can be used to show that the statement holds for the other sides.