Area of Triangle in Terms of Side and Altitude

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

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

Corollary
The area of a triangle $ABC$ is given by:
 * $\displaystyle \frac {a b \sin C} 2 = \frac {a c \sin B} 2 = \frac {b c \sin A} 2$

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.

Proof of Corollary
Follows from the (geometric) definition of sine.

Using the notation of the above triangle:

Substituting in:
 * $\displaystyle \frac {c \cdot h_c} 2 = \frac {b \cdot h_b} 2 = \frac {a \cdot h_a} 2$

... gives:
 * $\displaystyle \frac {a b \sin C} 2 = \frac {a c \sin B} 2 = \frac {b c \sin A} 2$

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.