# Right Triangle Similarity

## Big Picture

Right triangles have a lot of special characteristics, and similarity is no exception. Understanding similar right triangles is very useful in geometry and trignometry.

## Key Terms

Altitude: A line segment drawn perpendicularly from a vertex of the triangle to the other side.

Geometric Mean: The nth root of the product of n numbers.

## Right Triangle Altitudes

An altitude is a perpendicular segment that connects the vertex of a triangle to the opposite side. It is also known as the height of the triangle. The altitude of right triangles has a special attribute.

Theorem: If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other. Since the triangles are similar, side proportions can be used to find the lengths of any missing lengths or altitude.

$$\frac{DA}{AB} = \frac{AC}{CB} = \frac{DC}{CA}$$

$$\frac{DA}{DB} = \frac{AC}{AB} = \frac{DC}{DA}$$

$$\frac{AB}{DB} = \frac{CB}{AB} = \frac{CA}{DA}$$

## Geometric Mean

The geometric mean of two positive numbers a and b is the number x so that $$\frac{a}{x} = \frac{x}{b}$$

$$\frac{a}{x} = \frac{x}{b} \implies x^2 = ab \implies x = \sqrt{ab}$$

The geometric mean can be used to FInd the altitude of a right triangle. In a right triangle, the altitude drawn from the right angle to the hypotenuse divides the hypotenuse into two segments. In the triangle above, the hypotenuse is divided into $$\color{red}{DC}\color{black}{\text{ and }}\color{blue}{CB}$$.

# Right Triangle Similarity Cont.

## Geometric Mean (cont.)

Theorem: The length of the altitude is the geometric mean of the two segments.

$$\frac{BC}{AC} = \frac{AC}{DC} \implies AC\sqrt{BC \cdot DC}$$, where AC is the altitude

Theorem: The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.

$$\frac{AB}{CB} = \frac{CB}{DB} \implies CB\sqrt{AB \cdot DB}$$, where CB is one of the legs

$$\frac{AB}{AC} = \frac{AC}{AD} \implies AC\sqrt{AB \cdot AD}$$, where AC is the other leg Redraw the three triangles side-by-side so that the corresponding angles are in the same relative positions. 