AAA Similarity

Theorem: If in two triangles, corresponding angles are equal, then the triangles are similar.

Prerequisites:
Similarity of triangles (definition)
Basic Proportionality Theorem (proof)
Alternate angle property (proof)
SAS congruence (proof)

Proof:

Let there be two triangles $\triangle ABC$ and $\triangle DEF$, such that $\angle A = \angle D$, $\angle B = \angle E$ and $\angle C = \angle F$.

We need to show that $\triangle ABC\sim\triangle DEF$. For this draw a line PQ, where P and Q lies on the lines DE and DF respectively, such that AB $=$ DP and AC $=$ DQ.

In $\triangle ABC$ and $\triangle DPQ$,

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

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

Therefore, by CPCTC,

$\;\qquad\quad\angle B = \angle DPQ$
But, $\;\quad\angle B = \angle E\qquad\qquad\qquad\qquad\text{(given)}$
Hence, $\;\angle DPQ = \angle E$

Thus, by alternate angle property, PQ $\parallel$ EF.

Now using Basic Proportionality theorem in $\triangle DEF$,

$\qquad\quad\dfrac{PE}{DP} = \dfrac{QF}{DQ}\\[12pt] \Rightarrow\quad\;\;\dfrac{PE}{DP} + 1 = \dfrac{QF}{DQ} + 1\\[12pt] \Rightarrow\quad\;\;\dfrac{DP + PE}{DP} = \dfrac{DQ + QF}{DQ} \\[12pt] \Rightarrow\quad\;\;\dfrac{DE}{DP}= \dfrac{DF}{DQ}\\[12pt] \Rightarrow\quad\;\;\dfrac{DE}{AB} = \dfrac{DF}{AC}\qquad\qquad\qquad\qquad\text{(by construction)}$

Similarly,

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

Hence,

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

Thus, $\triangle ABC\sim\triangle DEF$  by the definition of similarity of triangles.


Recommended:
SSS similarity
Midpoint Theorem
Angle Bisector Theorem