SAS Similarity

Theorem: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

Prerequisites:
AA similarity (proof)
Converse of Basic Proportionality Theorem (proof)
SAS congruence (proof)
Corresponding angles property (proof)

Proof:

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

$\qquad\quad\dfrac{AB}{DE} = \dfrac{AC}{DF}\\[12pt]
\text{And, }\;\;\angle A = \angle D$

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.

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

$\qquad\quad AB = DP\qquad\qquad\qquad\text{(by construction)}\\
\qquad\quad AC = DQ\qquad\qquad\qquad\text{(by construction)}\\
\qquad\quad\angle A = \angle D\qquad\qquad\qquad\:\!\!\:\text{(given)}$

Hence, $\triangle ABC\cong\triangle DPQ$ by SAS rule of congruence.

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

Also, since $\angle D$ is common in $\triangle DPQ$ and $\triangle DEF$, hence by AA similarity,

$\qquad\quad\!\triangle DPQ\sim\triangle DEF\\
\therefore\quad\;\;\;\triangle ABC\sim\triangle DEF\qquad\qquad\qquad\text{(as $\triangle ABC\cong\triangle DPQ$)}$

Q.E.D.


Recommended:
SSS similarity
Midpoint Theorem
Pythagoras Theorem