The similarity theorems

Two triangles are similar when their angles are equal in pairs and their corresponding sides are in one ratio – the similarity ratio $\displaystyle k$.

• $\displaystyle \triangle ABC\sim\triangle DEF$ says which vertex goes with which: $\displaystyle A$ with $\displaystyle D$, $\displaystyle B$ with $\displaystyle E$, $\displaystyle C$ with $\displaystyle F$;
• so $\displaystyle {\angle A}={\angle D}$, $\displaystyle {\angle B}={\angle E}$, $\displaystyle {\angle C}={\angle F}$, and

• corresponding sides are opposite equal angles.

The AA similarity theorem: If two angles of one triangle are equal to two angles of another, the triangles are similar.

Given: $\displaystyle {\angle A}={\angle D}$ and $\displaystyle {\angle B}={\angle E}$ in the triangles $\displaystyle ABC$ and $\displaystyle DEF$.
To prove: $\displaystyle \triangle DEF\sim\triangle ABC$.

The proof below puts a copy of the triangle $\displaystyle DEF$ at the vertex $\displaystyle A$: $\displaystyle G$ on $\displaystyle AB$ with $\displaystyle {AG}={DE}$, and $\displaystyle H$ on $\displaystyle AC$ with $\displaystyle {AH}={DF}$. The copy has $\displaystyle GH\parallel BC$, and the extended Thales theorem gives the ratios.

Two more theorems prove that triangles are similar:

• SAS: if two sides of one triangle are proportional to two sides of another, and the angles between them are equal, the triangles are similar;
• SSS: if the three sides of one triangle are proportional to the three sides of another, the triangles are similar.

In a proof with similarity:

• find two triangles whose sides are the segments in the claim;
• match the vertices by the equal angles, and write the similarity in that order;
• read the proportion – or the equal angles – off the similarity.

In the triangle $\displaystyle ABC$, $\displaystyle D$ is on $\displaystyle AB$ and $\displaystyle E$ is on $\displaystyle AC$.
$\displaystyle {\angle AED}={\angle ABC}$.
$\displaystyle {AD}={4}$, $\displaystyle {AE}={6}$, $\displaystyle {AB}={12}$.
Find $\displaystyle AC$.

• $\displaystyle \angle A$ is common, and $\displaystyle {\angle AED}={\angle ABC}$, so $\displaystyle \triangle AED\sim\triangle ABC$ (AA);
• $\displaystyle E$ goes with $\displaystyle B$ and $\displaystyle D$ with $\displaystyle C$, so $\displaystyle {\frac{AE}{AB}}={\frac{AD}{AC}}$;
• $\displaystyle {\frac{6}{12}}={\frac{4}{AC}}$, so $\displaystyle {AC}={8}$.

$\displaystyle DE$ is not parallel to $\displaystyle BC$ here: the equal angles are at $\displaystyle E$ and at $\displaystyle B$, not at $\displaystyle D$ and at $\displaystyle B$.

In the triangle $\displaystyle ABC$, $\displaystyle {\angle BAC}={90^{\circ}}$.
$\displaystyle AD$ is the altitude to the hypotenuse $\displaystyle BC$.
Prove: $\displaystyle \triangle ABD\sim\triangle CAD$.

In the same triangle, prove: $\displaystyle {AD\cdot AD}={BD\cdot DC}$.
(A hint: Exercise 1.)

In a right-angled triangle, the altitude to the hypotenuse divides it into segments of $\displaystyle 4$ and $\displaystyle 9$.
Find the length of the altitude.

In the triangle $\displaystyle ABC$, $\displaystyle D$ is on the side $\displaystyle AB$ and $\displaystyle E$ is on the side $\displaystyle AC$.
$\displaystyle {AD}={6}$, $\displaystyle {AE}={5}$, $\displaystyle {AB}={10}$ and $\displaystyle {AC}={12}$.
Prove: $\displaystyle \triangle ADE\sim\triangle ACB$.

Are the triangles similar? If they are – what is the similarity ratio?

1. The sides of one triangle are $\displaystyle 4$, $\displaystyle 6$, $\displaystyle 8$, and of the other $\displaystyle 6$, $\displaystyle 9$, $\displaystyle 12$.
2. The sides of one triangle are $\displaystyle 4$, $\displaystyle 6$, $\displaystyle 8$, and of the other $\displaystyle 5$, $\displaystyle 7$, $\displaystyle 9$.