# Year 10+ Plane Geometry

### Chapters

### Theorems About Similar Triangles

# 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.

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

Find the length of \(CE\).

**Solution:**

By the Triangle Proportionality Theorem, \(\dfrac{AB}{BD} = \dfrac{AC}{CE}\). So,

## 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

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

Find the length of \(AC\).

**Solution:**

By the Angle Bisector Theorem, \(\dfrac{AB}{BD} = \dfrac{AC}{DC}\). So,

## 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

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

\(\dfrac{y}{b} = dfrac{p}{q}\) 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

### Example 3

The area of triangle \(ABC\) is \(60 \text{ cm}^2\). If triangles \(ABC\) and \(XYZ\) are similar, what is the area of triangle \(XYZ\)?

**Solution:**

The two triangles are similar, with corresponding sides in the ratio \(6: 9\), so their areas are in the ratio \(6^2:9^2 = 36:81 = 4:9\). Therefore, the area of triangle \(XYZ\) is equal to

### Description

In these chapters you will learn about plane geometry topics such as

- Area (Irregular polygons, plane shapes etc)
- Perimeter
- Conic sections (Circle, Ellipse, Hyperbola etc)
- Polygons (Congruent, polygons, similar, triangles etc)
- Transformations and symmetry (Reflection, symmetry, transformations etc)

etc

Even though these chapters are marked for Year 10 or higher students, several topics are for students in Year 8 or higher

### Audience

Year 10 or higher, suitable for Year 8 + students as well.

### Learning Objectives

Learn about Plane Geometry

Author: Subject Coach

Added on: 28th Sep 2018

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