Area of Triangle in Terms of Side and Altitude

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

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

Proof

 * Area-of-Triangle.png

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

From Halves of Parallelogram Are Congruent Triangles:
 * $\triangle ABC \cong \triangle DCB$

hence their areas are equal.

The Area of Parallelogram is equal to the product of one of its bases and the associated altitude.

Thus

where $\paren {XYZ}$ 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.