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Year 10+ Plane Geometry


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 \(\dfrac{AB}{BD} = \dfrac{AC}{CE}\).

Theorems About Similar Triangles

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:

Theorems About Similar Triangles

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

Theorems About Similar Triangles

Find the length of \(CE\).

Solution:

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

\( \begin{align*} \dfrac{4}{1} &= \dfrac{3}{CE}\\ CE &= \dfrac{3}{4} \end{align*} \)

The Angle Bisector Theorem

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

Theorems About Similar Triangles

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

Theorems About Similar Triangles

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

    \(\dfrac{\sin x^\circ}{BD} = \dfrac{\sin y^\circ}{AB}\).
    Rearranging yields
    \(\dfrac{AB}{BD} = \dfrac{\sin y^\circ}{\sin x^\circ}\).
    Similarly, applying the Sine Law to \(\triangle ACD\) gives
    \(\dfrac{AC}{DC} = \dfrac{\sin (180 - y)^\circ}{\sin x^\circ}\).

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

    \( \dfrac{AC}{DC} = \dfrac{\sin y^\circ}{\sin x^\circ} = \dfrac{AB}{BD}. \)

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

    Theorems About Similar Triangles

    Example 2

    Theorems About Similar Triangles

    Find the length of \(AC\).

    Solution:

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

    \( \begin{align*} \dfrac{9}{4} &= \dfrac{AC}{6}\\ AC&= \dfrac{54}{4} = 13.5 \end{align*} \)

    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 \(p^2:q^2\)

    Theorems About Similar Triangles

    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

    \( \begin{align*} \text{Area}_{XYZ} &= \dfrac{1}{2}\;yz \sin X\\ &= \dfrac{1}{2}\;\left(\dfrac{bp}{q}\right) \left(\dfrac{cp}{q}\right) \sin X\\ &= \dfrac{1}{2} \; \dfrac{p^2}{q^2} bc \sin A\\ &= \dfrac{p^2}{q^2} \text{Area}_{ABC} \end{align*} \)
    So, the areas of the two triangles are in the ratio:
    \( \text{Area of triangle }ABC: \text{Area of triangle }XYZ = p^2:q^2, \)
    as required.

    Example 3

    Theorems About Similar Triangles

    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

    \( \dfrac{4}{9}\text{Area}_{ABC} = \dfrac{4}{9} (60) = 26.6666\dots \text{ cm}^2. \)

    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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