Two triangles are said to be similar when they have their respective lengths proportional and their respective angles equal.
This implies a ratio \(k\) between the respective lengths, corresponding to an enlargement, a shrinkage or a preservation (if \(k = 1\)).
$$ \exists! k \in \mathbb{R}, \ \begin{Bmatrix} \overline{A'B'} = k \times \overline{AB}\\ \overline{B'C'} = k \times \overline{BC} \\ \overline{A'C'} = k \times \overline{AC} \end{Bmatrix} $$
$$ \begin{Bmatrix} \widehat{BAC} = \widehat{B'A'C'} = \alpha \\ \widehat{ABC} = \widehat{A'B'C'} = \beta \\ \widehat{BCA}= \widehat{B'C'A'} = \gamma \end{Bmatrix} $$
There are mainly three cases of triangle similarities.
Case 1: Three respective proportional sides
Two triangles are similar if their three respective sides are proportional.
Case 2: Two pairwise equal angles
Two triangles are similar if they have at least two pairwise equal angles respectively.
Case 3: A common angle and two proportional lengths
Two triangles are similar if they have a common angle and their two respective lengths are proportional.
Let us consider a triangle \(ABC\) and its image \(A'B'C'\) with the three respective proportional sides.
$$ \exists! k \in \mathbb{R}, \ \begin{Bmatrix} \overline{AB} = k \times \overline{AB'}\\ \overline{BC} = k \times \overline{BC'} \\ \overline{AC} = k \times \overline{AC'} \end{Bmatrix} $$
The triangle \(A'B'C'\) is therefore an enlargement (or a shrinkage depending on \(k\)) of the starting triangle \(ABC\).
The angles are also well preserved, this can be verified on the angle \( \beta \) because:
$$ \begin{Bmatrix} cos(\beta) = \frac{AH}{AB}\\ cos(\beta') = \frac{A'H'}{A'B'} = \frac{k \times AH }{k \times AB} = \frac{AH}{AB} \end{Bmatrix} \Longrightarrow cos(\beta) = cos(\beta') \qquad (1) $$
$$ \begin{Bmatrix} sin(\beta) = \frac{BH}{AB}\\ sin(\beta') = \frac{B'H'}{A'B'} = \frac{k \times BH }{k \times AB} = \frac{BH}{AB} \end{Bmatrix} \Longrightarrow sin(\beta) = sin(\beta') \qquad (2) $$
Thanks to both expressions \( (1)\) and \( (2)\), we can see that:
$$\beta = \beta' $$
By repeating the same process for the angles \( \alpha \) and \( \gamma \) respectively corresponding to the vertices \( A \) and \( C \), we can assert that:
$$ \begin{Bmatrix} \alpha = \alpha' \\ \beta = \beta' \\ \gamma = \gamma' \end{Bmatrix} $$
These two triangles have the same respective angle measurements.
And finally,
Two triangles are similar if their three respective sides are proportional.
Let us consider two nested triangles \(ABC \) and \(ADE \), having two angles in common:
- \(\alpha\) corresponding to the common vertex \(A\)
- \(\beta\) corresponding to the vertices \(B\) and \(D\)
Having two pairwise equal angles respectively, it is obvious that the third is also in common because the sum of the angles of a triangle always equals \( \pi\). The third angle \( \gamma \) is therefore equal to:
$$ \gamma = \pi - (\alpha + \beta) $$
Thus, we do have the following equations:
$$ \begin{Bmatrix} \widehat{BAC} = \widehat{DAE} = \alpha \\ \widehat{ABC} = \widehat{ADE} = \beta \\ \widehat{BCA}= \widehat{DEA} = \pi - (\alpha + \beta) \end{Bmatrix} $$
Furthermore, the lengths \(BC \) and \(DE \) intersecting the same line \((AB) \) with an equal angle, we do have that \( (BC) \parallel (DE) \).
We can therefore apply Thales' theorem. We do have the following relations:
$$ \exists! k \in \mathbb{R}, \ \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} = k $$
$$ \begin{Bmatrix} \overline{AD} = k \times \overline{AB}\\ \overline{AE} = k \times \overline{AC} \\ \overline{DE} = k \times \overline{BC} \end{Bmatrix} $$
We definitely have the three proportional lengths as well as the respective equal angles.
Then finally,
Two triangles are similar if they have at least two pairwise equal angles respectively.
Let us consider two nested triangles \(ABC \) and \(ADE \), having a common vertex \(A\) with a corresponding common angle \(\alpha\), as well as two respective sides of the same proportion starting from this vextex, and such as:
$$ \exists! k \in \mathbb{R}, \ \begin{Bmatrix} \overline{AD} = k \times \overline{AB}\\ \overline{AE} = k \times \overline{AC} \\ \widehat{BAC} = \widehat{DAE} = \alpha \end{Bmatrix} $$
Then,
$$ \exists! k \in \mathbb{R}, \ \frac{AD}{AB} = \frac{AE}{AC} = k \qquad (3) $$
Thanks to The Al-Kashi's theorem, we know that:
In a context of an ordinary triangle \(\{a, b, c\}\), which every angle \(\alpha, \beta, \gamma \) in front of its correspondant side, such as:
$$ \left \{ \begin{gather*} \alpha \enspace opposed \enspace to \enspace a \\ \beta \enspace opposed \enspace to \enspace b \\ \gamma \enspace opposed \enspace to \enspace c \end{gather*} \right \} $$
and such as the following figure:
We do have the three follolwing relations:
$$ a^2 = b^2 + c^2 - 2bc.cos(\alpha) \qquad (Al-Kashi) $$
$$ b^2 = a^2 + c^2 - 2ac.cos(\beta) \qquad (Al-Kashi^*) $$
$$ c^2 = a^2 + b^2 - 2ab.cos(\gamma) \qquad (Al-Kashi^{**}) $$
Then, in our case, we do have:
$$DE^2 = AD^2 + AE^2 - 2 AD.AE.cos(\vec{AD}, \vec{AE}) \qquad (4) $$
But also:
$$BC^2 = AB^2 + AC^2 - 2 AB.AC.cos(\vec{AB}, \vec{AC}) $$
Now, replacing the values of \(AB\) and \(AC\) by their proportional value according to \((3)\):
$$BC^2 = \left(\frac{AD}{k} \right)^2 + \left(\frac{AE}{k} \right)^2 - 2 \left(\frac{AD}{k} \right).\left(\frac{AE}{k} \right).cos(\vec{AD}, \vec{AE}) $$
$$BC^2 = \left(\frac{AD^2}{k^2} \right) + \left(\frac{AE^2}{k^2} \right) - \frac{2}{k^2}. AD.AE.cos(\vec{AD}, \vec{AE}) $$
$$k^2 BC^2 = AD^2 + AE^2 - 2 AD.AE.cos(\vec{AD}, \vec{AE}) \qquad (5) $$
And injecting \((5) \) into \((4) \), we finally have that:
$$k^2 BC^2 = DE^2 \Longrightarrow k = \frac{DE}{BC} \qquad (5) $$
Thanks to \((6)\), we obtain a third equality added to \((3)\):
$$ \exists! k \in \mathbb{R}, \ \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} = k $$
Furthermore, we saw with the first case of similarities above that this also implied the preservation of the respective angles.
Thus,
Two triangles are similar if they have a common angle and their two respective lengths are proportional.
It is possible to apply hte same reasoning even if the common angle is not the common vertex, and this leads us to the same result anyway.
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