Today, we are going to learn about

*similar triangles*. We will use similar triangles to give a proof of Pythagorean Theorem.

As suggested by the name, two similar triangles simply look similar as in the figure below

Two similar triangles have the following two important properties:

**Three pairs of equal angles**

$$ \angle A = \angle A', ~~~

\angle B = \angle B', ~~~

\angle C = \angle C' $$

**Three pairs of sides have equal ratios**

$$ \frac{AB}{A'B'} =

\frac{BC}{B'C'} =

\frac{CA}{C'A'} $$

So how do we prove two given triangles are similar? There are three common ways.

**Angle-Angle Case:**

**If two triangles have two pairs of equal angles then they are two similar triangles**

In the figure below, if it can be shown that $$\angle A = \angle A' ~~~\mathrm{ and }~~~ \angle B = \angle B'$$ then we can conclude that the two triangles $ABC$ and $A'B'C'$ are similar.

**Side-Side-Side Case:**

**If two triangles have three pairs of sides of equal ratios then they are two similar triangles**

In the figure below, if it can be shown that

$$ \frac{AB}{A'B'} =

\frac{BC}{B'C'} =

\frac{CA}{C'A'} $$

then we can conclude that the two triangles $ABC$ and $A'B'C'$ are similar.

**Side-Angle-Side Case:**

**If two triangles have two pairs of sides of equal ratios and the two angles between these two side pairs are equal then they are two similar triangles**

In the figure below, if it can be shown that $$\frac{AB}{A' B'} = \frac{BC}{B' C'} ~~~\mathrm{ and }~~~ \angle B = \angle B'$$ then we can conclude that the two triangles $ABC$ and $A'B'C'$ are similar.

If the two given triangles are

**right-angled**triangles then it is even simpler to prove that they are similar. We have the following cases.

**Acute Angle Case:**

**If two**

*right-angled*triangles have one pair of equal acute angles then they are similar trianglesIn the figure below, if it can be shown that $$\angle A = \angle A'$$ then we can conclude that the two right-angled triangles $ABC$ and $A'B'C'$ are similar.

**Side-Side Case:**

**If two**

*right-angled***triangles have two pairs of sides of equal ratios then they are two similar triangles**

In the above figure, if it can be shown that either $$\frac{AB}{A' B'} = \frac{BC}{B' C'}, ~~~\mathrm{ or }~~~ \frac{BC}{B' C'} = \frac{CA}{C' A'}, ~~~\mathrm{ or }~~~ \frac{CA}{C' A'} = \frac{AB}{A' B'}$$ then we can conclude that the two right-angled triangles $ABC$ and $A'B'C'$ are similar.

Now we are going to use similar triangles to prove a basic geometry theorem -- the Pythagorean theorem.

**Pythagorean theorem**

Given a triangle $ABC$ with a right angle $A$. Prove that

$$ BC^2 = AB^2 + AC^2. $$

**Proof:**Draw the altitude $AH$ onto the side $BC$.

Consider the two right-angled triangles $ABC$ and $HBA$. These two right-angled triangles have a pair of equal angles at the vertex $B$, thus, they are similar triangles. It follows that $$\frac{AB}{HB} = \frac{BC}{BA},$$ and hence, $$AB^2 = HB \times BC.$$

Similarly, consider the two right-angled triangles $ABC$ and $HAC$. These two right-angled triangles have a pair of equal angles at the vertex $C$, thus, they are similar triangles. It follows that $$\frac{AC}{HC} = \frac{BC}{AC},$$ and hence, $$AC^2 = HC \times BC.$$

Therefore, we have

$$ AB^2 + AC^2 = HB \times BC + HC \times BC = BC^2. \blacksquare $$

You can find a proof of Morley theorem that uses similar triangles here http://mathgardenblog.blogspot.com/2012/05/morley-theorem.html.

*Homework:*Prove the

*angle bisector theorem*below

Given a triangle ABC. Draw the bisector AD of the angle A. Prove that

$$

\frac{AB}{AC} = \frac{DB}{DC}.

$$

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