Area of Triangle in Terms of Side and Altitude

Theorem
The area of a triangle $$ABC\,$$ is given by:
 * $$\frac {c \cdot h_c} {2} = \frac {b \cdot h_b} {2} = \frac {a \cdot h_a} {2}$$

where:
 * $$a, b, c$$ are the sides, and
 * $$h_a, h_b, h_c$$ are the altitudes from $$A$$, $$B$$ and $$C$$ respectively.

Proof

 * [[File:Tri.PNG]]

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

$$ $$ $$

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.

Note
This formula is perhaps the best-known and most useful for determining a triangle's area.

It is usually remembered, and quoted, as "half base times height".