similarity in triangles

Similarities between figures

Two figures are said to be similar if they have the same shape but not necessarily the same size. The following figures show similar circles and similar triangles.

Similarity of Triangles: Two triangles which have the three angles of one triangle equal respectively to the three angles of other triangle and all the ratios between the measure of corresponding sides equal are said to be similar. 

According to the definition, triangle ABC is similar to triangle PQR , \(\triangle ABC \sim \triangle PQR\) if:

1. \(\displaystyle \frac{a}{p} = \frac{b}{q} = \frac{c}{r}\)
2. \(\displaystyle \angle A = \angle P, \angle B = \angle Q, \angle C = \angle R\)

So, the similarity of triangles requires two things:

• Corresponding angles must be equal 
• Corresponding sides must be proportional

THREE TESTS FOR SIMILARITY OF TRIANGLES

1.  Angle-Angle-Angle (\(AAA\)) AXIOM OF SIMILARITY

If two triangles have two pairs of angles equal, their corresponding sides are proportional. In triangle ABC and \(DEF\), \(\displaystyle \angle A = \angle D, \angle B = \angle E, \angle C = \angle F\), then \(\triangle ABC \sim \triangle DEF\) , that is 
\(\displaystyle \frac{BC}{EF} = \frac{AB}{DE}= \frac{AC}{DF}\)

2. Side-Angle-Side (\(SAS\)) AXIOM OF SIMILARITY

If two triangles have a pair of corresponding angles equal and sides including them proportional then the triangles are similar. 

If in triangle ABC and \(DEF\), \(\angle A = \angle D\) and \(\frac{AB} { DE} = \frac{AC}{DF}\)  then  \(\triangle ABC \sim \triangle DEF\)

3. Side-Side-Side (\(SSS\)) AXIOM OF SIMILARITY

If two triangles have their pairs of corresponding sides proportional then the triangles are similar. If two triangles ABC and \(DEF\), \(\displaystyle \frac{BC}{EF} = \frac{AB}{DE}= \frac{AC}{DF}\) then \(\triangle ABC \sim \triangle DEF\)

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Theorem 1

A straight line drawn parallel to one side of a triangle divides the other two sides proportionally. Conversely, if a line divides any two sides of a triangle proportionally then the line is parallel to the third side.

In \(\triangle ABC, \ DE \parallel BC\) then

\(\displaystyle \frac{AD}{DB} = \frac{AE}{EC}\)

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Theorem 2

The areas of similar triangles are proportional to the squares on corresponding sides. 

\(\displaystyle \frac{\textrm{area of }\triangle ABC}{\textrm{area of }\triangle DEF} = \) \(\displaystyle \frac{BC^2}{EF^2} = \frac{AB^2}{DE^2}= \frac{AC^2}{DF^2}\)

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Example 1: In \(\triangle ABC, PQ \parallel BC\). If AP/\(PB\) = 1/2 and AQ = 2 cm. Find QC.

As PQ is parallel to BC, therefore
\(\displaystyle \frac{AP}{PB} = \frac{AQ}{QC}\)

\(\displaystyle\frac{1}{2} = \frac{2}{QC}\)⇒ \(\displaystyle QC = 2 \times 2 = 4\)

Example 2: Triangle \(\triangle ABD, \triangle ACD \) are similar. BD = 2 cm and AB = 3 cm. If the area of the triangle \(\triangle ABD \) is 2 cm², calculate area of \( \triangle ACD \).

\(\displaystyle \frac{\textrm{area of }\triangle ABD}{\textrm{area of }\triangle ADC} = \frac{4}{DC^2} = \frac{9}{AC^2}\)

\( \frac{2}{\textrm{Area of }\triangle ADC} = \frac{4}{9}\)

\(\textrm{Area of }\triangle ADC = \frac{2 \times 9}{4} = 4.5 cm^2\)