Theorems About Similar Triangles

The Triangle Proportionality Theorem

This theorem states that if \(ADE\) is a triangle, and \(BC\) is drawn parallel to the side \(DE\), then \(\dfrac{AB}{BD} = \dfrac{AC}{CE}\).

Proof: To prove this theorem (show it is true), we add an extra line \(BF\) parallel to \(AE\) to form parallelogram \(BCEF\) as shown in the diagram below:

Next, we show that triangles \(ABC\) and \(BDF\) are similar:

• \(\angle ABC = \angle BDF\) as they are corresponding angles.
• \(\angle BFD = \angle ACB\) as they are alternate (interior) angles.

Therefore, the two triangles are similar as they are equiangular.

We use corresponding sides of similar triangles \(ABC\) and \(BDF\) to establish the result:

• Sides \(AB\) and \(BD\), and sides \(AC\) and \(BF\) are corresponding sides.
• Since the triangles are similar, \(\dfrac{AB}{BD} = \dfrac{AC}{BF}\).
• \(BF = CE\) (opposite sides of parallelogram \(BCEF\)).
• So, \(\dfrac{AB}{BD} = \dfrac{AC}{CE}\), establishing our result.

Example 1

The Angle Bisector Theorem

This theorem states that if \(ABC\) is a triangle, and \(AD\) is drawn so that it bisects \(\angle BAC\), then \(\dfrac{AB}{BD} = \dfrac{AC}{DC}\).

Proof: To prove this theorem (show it is true), we label the triangle as as shown in the diagram below:

Then

\(\angle BAD = \angle DAC = x^\circ\) as \(AD\) bisects \(\angle BAC\).

and we let \(\angle ADB = y^\circ\), so, as they are angles on the same straight line, \(\angle ADC = (180 - y)^\circ\).

Using the Sine Law on \(\triangle ACD\) gives

Rearranging yields Similarly, applying the Sine Law to \(\triangle ACD\) gives

The result follows since \(\sin (180 - y)^\circ = \sin y^\circ\). Therefore,

An interesting consequence of this result is that if \(\triangle ABC\) is isosceles, then triangles \(ABD\) and \(ACD\) are congruent.

Example 2

Areas of Similar Triangles

If triangles \(ABC\) and \(XYZ\) are similar and have sides in the ratio \(p:q\), then their areas are in the ratio \(p^2:q^2\)

Proof

Suppose triangles \(ABC\) and \(XYZ\) are similar and have sides in the ratio \(p:q\). Then we can use the general formula for the area of a triangle to find

• \(\text{Area}_{ABC} = \dfrac{1}{2}\;bc \sin A\)
• \(\text{Area}_{XYZ} = \dfrac{1}{2}\;yz \sin X\)

Since these lengths are in the ratio \(p:q\),
\(\dfrac{y}{b} = dfrac{p}{q}\) and \(\dfrac{z}{c} = dfrac{p}{q}\).

So, \(y = \dfrac{bp}{q}\) and \(z = \dfrac{cp}{q}\).

We also have \(A = X\) since the triangles are similar.

So, we can calculate

So, the areas of the two triangles are in the ratio: as required.

Example 3