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:
$$ \angle A = \angle A', ~~~
\angle B = \angle B', ~~~
\angle C = \angle C' $$
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 triangles
In 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
$$
\frac{AB}{AC} = \frac{DB}{DC}.
$$
