# Theorems about Similar Triangles

## 1. The Side-Splitter Theorem

If ADE is any triangle and BC is drawn parallel to DE, then \frac{AB}{BD} = \frac{AC}{CE}

To show this is true, draw the line BF parallel to AE to complete a parallelogram BCEF:

Triangles ABC and BDF have exactly the same angles and so are similar (Why? See the section called ** AA** on the page How To Find if Triangles are Similar.)

- Side AB corresponds to side BD and side AC corresponds to side BF.
- So AB/BD = AC/BF
- But BF = CE
- So AB/BD = AC/CE

## The Angle Bisector Theorem

If ABC is any triangle and AD bisects (cuts in half) the angle BAC, then \frac{AB}{BD} = \frac{AC}{DC}

To show this is true, we can label the triangle like this:

- Angle BAD = Angle DAC = x°
- Angle ADB = y°
- Angle ADC = (180−y)°

**\frac{AB}{BD} = \frac{sin(y)}{sin(x)}**

**sin(180−y) = sin(y)**:

**\frac{AC}{DC} = \frac{sin(y)}{sin(x)}**

Both \frac{AB}{BD} and \frac{AC}{DC} are equal to \frac{sin(y)}{sin(x)}, so:

\frac{AB}{BD} = \frac{AC}{DC}

In particular, if triangle ABC is isosceles, then triangles ABD and ACD are congruent triangles

And the same result is true:

\frac{AB}{BD} = \frac{AC}{DC}

## 3. Area and Similarity

If two similar triangles have sides in the ratio x:y,

then their areas are in the ratio x^{2}:y^{2}

### Example:

These two triangles are similar with sides in the ratio 2:1 (the sides of one are twice as long as the other):

What can we say about their areas?

The answer is simple if we just draw in three more lines:

We can see that the small triangle fits into the big triangle **four times**.

So when the lengths are **twice** as long, the area is **four times** as big

So the ratio of their areas is 4:1

We can also write 4:1 as 2^{2}:1

### The General Case:

Triangles ABC and PQR are similar and have sides in the ratio **x:y**

We can find the areas using this formula from Area of a Triangle:

Area of ABC = \frac{1}{2}bc sin(A)

Area of PQR = \frac{1}{2}qr sin(P)

And we know the lengths of the triangles are in the ratio **x:y**

q/b = y/x, so: **q = by/x**

and r/c = y/x, so **r = cy/x**

Also, since the triangles are similar, **angles A and P** are the same:

**A = P**

We can now do some calculations:

**Area of triangle PQR**:

**\frac{1}{2}qr sin(P)**

**Area of Triangle ABC**

So we end up with this ratio:

Area of triangle ABC : Area of triangle PQR = x^{2 }: y^{2}