SSS Similarity

Theorem: If the corresponding sides of two triangles are proportional, then they are similar.

Prerequisites:
Similarity of triangles (definition)
AA Similarity (proof)
Converse of Basic Proportionality Theorem (proof)
SSS congruence (proof)
Corresponding angles property (proof)

Proof:

Let there be two triangles $\triangle ABC$ and $\triangle DEF$, such that:

$\qquad\quad\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$

In order to prove that $\triangle ABC\sim\triangle DEF$, let us draw a line PQ, where P lies on the line DE and Q lies on the line DF, such that DP $=$ AB and DQ $=$ AC.

$\text{Since, }\;\dfrac{AB}{DE} = \dfrac{AC}{DF}\\[12pt]
\text{Hence, }\dfrac{DP}{DE} = \dfrac{DQ}{DF}\qquad\qquad\qquad\text{(by construction)}\\[12pt]
\therefore\quad\;\;\;\;\: PQ \parallel EF\qquad\qquad\qquad\;\quad\text{(by converse of Basic Proportionality Theorem)}\\
\Rightarrow\quad\;\;\:\angle DPQ = \angle E\qquad\qquad\qquad\!\!\:\!\text{(corresponding angles)}$

Also, since $\angle D$ is common in $\triangle DPQ$ and $\triangle DEF$,  thus,  $\triangle DPQ\sim\triangle DEF\;$ by AA similarity.

Therefore, by the definition of similarity of triangles,

$\qquad\quad\dfrac{DP}{DE} = \dfrac{PQ}{EF}\\[12pt]
\Rightarrow\quad\;\;\dfrac{AB}{DE} = \dfrac{PQ}{EF}\qquad\qquad\qquad\text{(by construction)}\qquad\cdots\text{(1)}\\[12pt]
\text{But, }\;\;\dfrac{AB}{DE} = \dfrac{BC}{EF}\qquad\qquad\qquad\text{(given)}\qquad\qquad\qquad\;\:\cdots\text{(2)}\\[12pt]
\Rightarrow\quad\;\;\:\!\dfrac{PQ}{EF}=\dfrac{BC}{EF}\qquad\qquad\qquad\!\!\:\text{(from $(1)$ and $(2)$)}\\[12pt]
\Rightarrow\quad\;\;\; PQ = BC\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\cdots\text{(3)}$

Now, in $\triangle ABC$ and $\triangle DPQ$,

$\qquad\quad AB = DP\qquad\qquad\qquad\text{(by construction)}\\
\qquad\quad AC = DQ\qquad\qquad\qquad\:\!\!\text{(by construction)}\\
\qquad\quad BC = PQ\qquad\qquad\qquad\text{(from $(3)$)}$

Hence, $\triangle ABC\cong\triangle DPQ$ by SSS rule.

Thus, since

$\qquad\quad\!\triangle DPQ\sim\triangle DEF\\
\therefore\quad\;\;\;\triangle ABC\sim\triangle DEF$

Q.E.D.


Recommended:
SAS similarity
Midpoint Theorem
Pythagoras Theorem