The Thales' theorem and its reciprocal

In an triangle, The Thales' theorem implies proportionality ratios between sides length.

Let be an ordinary triangle, in which we have drew a parallel of one of the sides, and such as the following figure:

The Thales' theorem

The Thales' theorem tells us that, in a triangle \(ABD\), if it exists any straight line \(BC\) intersecting \(AD\) and \(AE\) respectively on \(B\) and \(C\) such as \(BC \parallel DE\), then it implies lengths ratios between sides:

$$ BC \parallel DE \Longrightarrow \Biggl( \frac{AB}{AD} = \frac{AC}{AE} = \frac{BC}{DE} \Biggr) \qquad \bigl(Thales \enspace(theoreme) \bigr) $$

As well, it is possible to apply it in a reversed triangle, such as the following figure:

1. Extension of the theorem

Finally, by extension of the Thales theorem, if we have established these following equalities between ratios:

$$ \frac{AB}{AD} = \frac{AB'}{AD'} = \frac{BB'}{DD'} $$

These ratios will apply for all projections (not necessarily orthogonal) on the side \( DD' \).



$$ \frac{AB}{AD} = \frac{AB_1}{AD_1} = \frac{AB_2}{AD_2}= \frac{AB'}{AD'} $$

The Thales' theorem reciprocal

The Thales' theorem reciprocal tells us that if in any triangle \( ADE \), with any straight line \( BC \) intersecting \( AD \) and \( AE \) respectively on \( B \) and \( C \), such as the following figure:

Then, it implies a parallelism relation:

$$ \Biggl( \frac{AB}{AD} = \frac{AC}{AE} = \frac{BC}{DE} \Biggr) \Longrightarrow BC \parallel DE \qquad \bigl(Thales \enspace(reciprocal) \bigr) $$

Furthermore, only one over three ratio equalities is enough to imply the parallelism, and:

$$ \Biggl( \frac{AB}{AD} = \frac{AC}{AE}\Biggr) \ ou \ \Biggl(\frac{AB}{AD} = \frac{BC}{DE} \Biggr) \ ou \ \Biggl(\frac{AC}{AE} = \frac{BC}{DE}\Biggr) \Longrightarrow BC \parallel DE \qquad \bigl(Thales \enspace(reciprocal^*) \bigr) $$

The Thales' theorem equivalence

Both previous implications then form the following equivalence:

$$ BC \parallel DE \Longleftrightarrow \Biggl( \frac{AB}{AD} = \frac{AC}{AE} = \frac{BC}{DE} \Biggr) \qquad \bigl(Thales (equivalence) \bigr) $$

---

Demonstration

Let be an ordinary triangle, in which we added a straight line \(BC\) intersecting \(AD\) and \(AE\) respectively on \(B\) and \(C\).

The Thales' theorem

Let us start from the hypothesis that:

To proove the truthfulness of the theorem, let us add to this triangle an height \( AG \) intesecting \( BC \) and \( DE \), respectively on \( F \) and \( G \).

We can now apply the laws of trigonometry in it.

1. Trigonometry relations in the left part of the triangle

For both triangles \( ADG \) and \( ABF \), we do have:

$$ cos(\alpha) = \frac{DG}{DA} = \frac{BF}{BA} $$

Then, we can say that:

$$ \frac{DG}{DA} = \frac{BF}{BA} $$

And also:

$$ \frac{AB}{AD} = \frac{BF}{DG} \qquad (1) $$



As well,

$$ sin(\alpha) = \frac{AG}{AD} = \frac{AF}{AB} $$
$$ \frac{AG}{AD} = \frac{AF}{AB} $$
$$ \frac{AB}{AD} = \frac{AF}{AG} \qquad (2) $$



Now, we notice a commun term \( \frac{AB}{AD} \) in both equations \( (1) \) and \( (2) \), that gives us a triple equality:

$$ \frac{AB}{AD} = \frac{BF}{DG} = \textcolor{#446e4f}{\frac{AF}{AG}} \qquad (3) $$

Let us now do the same thing in the right part of the triangle.




2. Trigonometry relations in the right part of the triangle

Repeating the same process as before, we find out a new triple equality:

$$ \frac{AC}{AE} = \frac{FC}{GE} = \textcolor{#446e4f}{\frac{AF}{AG}} \qquad (4) $$



3. General equality between ratios

Now, we notice that in both expressions \( (3) \) and \( (4) \), there is a common term \( \frac{AF}{AG} \), thus all these ratios are equals:

$$ \frac{AB}{AD} = \frac{AC}{AE} = \frac{BF}{DG} = \frac{FC}{GE} = \textcolor{#446e4f}{\frac{AF}{AG}} \qquad (5) $$

So far, we prooved that:

$$ \frac{AB}{AD} = \frac{AC}{AE} $$



It remains to be prooved that these two ratios are also equals to the third: \( \frac{BC}{DE} \).




We know from the property of addition of numerators and denominators of a fraction that:

$$ \forall (a, c) \in \hspace{0.05em} \mathbb{R}^2, \enspace (b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2, \enspace \ \Bigl \{ (b+d) \Bigr \} \ \neq 0, $$

$$ \frac{a}{b} = \frac{c}{d} = \frac{a+c}{b+d} $$

.

In our case, the previous property allows us to say that:

$$ \frac{BF}{DG} = \frac{FC}{GE} = \frac{BF +FC}{DG + DE} $$

And by the way:

$$ \frac{BF}{DG} = \textcolor{#446e4f}{\frac{FC}{GE}} = \frac{BC}{DE} \qquad (6) $$



As with the group of equalities \( (5) \), we had:

$$ \frac{AB}{AD} = \frac{AC}{AE} = \frac{BF}{DG} = \textcolor{#446e4f}{\frac{FC}{GE}} = \frac{AF}{AG} \qquad (5) $$

The group of equalities \( (5) \) and \( (6) \) having at least one common term which is \( \frac{FC}{GE} \), then the ratio \( \frac{BC}{DE} \) is also equal to the others, and:

$$ \frac{AB}{AD} = \frac{AC}{AE} = \frac{BF}{DG} = \frac{FC}{GE} = \frac{AF}{AG} = \frac{BC}{DE} \qquad (5') $$

We thus prooved that:

$$ \frac{AB}{AD} = \frac{AC}{AE} = \frac{BC}{DE} $$




And as a result,

$$ BC \parallel DE \Longrightarrow \Biggl( \frac{AB}{AD} = \frac{AC}{AE} = \frac{BC}{DE} \Biggr) \qquad \bigl(Thales \enspace(theoreme) \bigr) $$


4. Extension of the theorem

Finally, by extension of the Thales theorem, if we have established these following equalities between ratios:

$$ \frac{AB}{AD} = \frac{AB'}{AD'} = \frac{BB'}{DD'} $$

These ratios will apply for all projections (not necessarily orthogonal) on the side \( DD' \).



$$ \frac{AB}{AD} = \frac{AB_1}{AD_1} = \frac{AB_2}{AD_2}= \frac{AB'}{AD'} $$

The Thales' theorem reciprocal

To now proove the veracity of the theorem reciprocal, let us start from three different hypotheses:

• \( \frac{AB}{AD} = \frac{AC}{AE} \qquad (H_2) \)
• \( \frac{AB}{AD} = \frac{BC}{DE} \qquad (H_2') \)
• \( \frac{AC}{AE} = \frac{BC}{DE} \qquad (H_2'') \)

1. En utilisant la similarité des deux triangles

We know from the following property of the similarity of two triangles that:




Two triangles are similar if they have a common angle and their two respective lengths are proportional.

As a consequence of it, with the three hypotheses \((H_2)\), \((H_2')\) and \((H_2'')\) we will always have similar triangles. And these two triangles being nested in one another, it is obvious to conclude that in any case:

$$ BC \parallel DE $$

2. Conclusion

We showed that one over three ratio equality was enough to imply parallelism, and:

$$ \Biggl( \frac{AB}{AD} = \frac{AC}{AE}\Biggr) \ ou \ \Biggl(\frac{AB}{AD} = \frac{BC}{DE} \Biggr) \ ou \ \Biggl(\frac{AC}{AE} = \frac{BC}{DE}\Biggr) \Longrightarrow BC \parallel DE \qquad \bigl(Thales \enspace(reciprocal^*) \bigr) $$

And if the three ratios are equal, we can write that:

$$ \Biggl( \frac{AB}{AD} = \frac{AC}{AE} = \frac{BC}{DE} \Biggr) \Longrightarrow BC \parallel DE \qquad \bigl(Thales \enspace(reciprocal) \bigr) $$

The Thales' theorem equivalence

Two implications makes an equivalence.

Thus, having our two implications \((I_1)\) et \((I_2)\):

$$ \left \{ \begin{align*} BC \parallel DE \Longrightarrow \Biggl( \frac{AB}{AD} = \frac{AC}{AE} = \frac{BC}{DE} \Biggr) \qquad (I_1) \\ \\ \Biggl( \frac{AB}{AD} = \frac{AC}{AE} = \frac{BC}{DE} \Biggr) \Longrightarrow BC \parallel DE \qquad (I_2) \end{align*} \right \} $$

We can gather them to build the following equivalence:

$$ BC \parallel DE \Longleftrightarrow \Biggl( \frac{AB}{AD} = \frac{AC}{AE} = \frac{BC}{DE}\Biggr) \qquad \bigl(Thales \enspace (equivalence) \bigr) $$