# Math Help - Proofs for similarity theorems (AA, SAS, SSS)

1. ## Proofs for similarity theorems (AA, SAS, SSS)

Hello forum! I am searching for proofs for the three theorems of similarity: AA (angle-angle), SAS (side-angle-side) and (side-side-side). I tried googling a bit and even checked YouTube but didn't find any (at least any that were satisfactorily produced). Does anyone got simple proofs for the theorems or perhaps some good links?

2. ## Re: Proofs for similarity theorems (AA, SAS, SSS)

We proceed by proving the following important result

Theorem: (Sine Rule) In a triangle $ABC$ $\frac{\sin \angle A}{a} = \frac{\sin \angle B}{b} = \frac{\sin \angle C}{c} = \frac{1}{2 R}$ where $BC = a, AC=b, AB=c$ and $R$ is the radius of circumcircle.

Proof:
In $\triangle BOC$ $\angle BOC = 2 \angle A, BO = R, \sin(\angle BOD) = \frac{\frac{a}{2}}{R} \implies \sin \angle A = \frac {a}{2R}$
Similarly in $\triangle COA, \triangle BOA$ we have $\sin \angle B = \frac {b}{2R}, \sin \angle C = \frac {c}{2R}$. Hence the proof.

Now for the similarity theorems we have start by assuming the AAA similarity postulate as the definition and prove SSS, SAS or assume SSS similarity postulate and prove AAA, SAS.

Here I assume SSS postulate and prove the other two theorems

AAA: Two triangles $\triangle ABC, \triangle A'B'C'$ such that $\angle A = \angle A', \angle B = \angle B', \angle C = \angle C'$ are similar.
From Sine Rule we have $\sin \angle A = \frac{a}{2R} = \sin \angle A' = \frac{a'}{2R'} \implies \frac{a}{a'} = \frac{R}{R'}$
Similarly $\frac{b}{b'} = \frac{R}{R'}, \frac{c}{c'} = \frac{R}{R'}$ hence by SSS we have $\triangle ABC \approx \triangle A'B'C'$

SAS: The argument is same as above use sine rule to get the sine angles are equal and argue that the angles are themselves equal and not supplementary. And use AAA.

Kalyan

3. ## Re: Proofs for similarity theorems (AA, SAS, SSS)

the angles are themselves equal and not supplementary. And use AAA.

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