# Area of Triangle in Terms of Side and Altitude/Proof 1

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## 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:

## Proof

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

From Opposite Sides and Angles of Parallelogram are Equal:

- $\triangle ABC \cong \triangle DCB$

hence their areas are equal.

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

Thus

\(\ds \paren {ABCD}\) | \(=\) | \(\ds c \cdot h_c\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds 2 \paren {ABC}\) | \(=\) | \(\ds c \cdot h_c\) | because congruent surfaces have equal areas | ||||||||||

\(\ds \paren {ABC}\) | \(=\) | \(\ds \frac {c \cdot h_c} 2\) |

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.

$\blacksquare$

## Motivation

**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**.