Case 1: Internal point of a circle
If a point \( O\) is the intersection between two ropes \( (AB )\) and \( (CD )\) of a circle inside it, and such as the following figure:
So, the two triangles \( AOC \) and \( DOB \) are similar, and the product between the lengths respectively proportional starting from the point \(O\) is worth:
$$\overline{OA} \times \hspace{0.05em} \overline{OB} =\hspace{0.05em} \overline{OC} \times \overline{OD}$$
Case 2: External point of a circle
If a point \( O\) is the intersection between two ropes \( (AC )\) and \( (DB )\) of a circle outside it, and such as the following figure:
So, the two triangles \( ABO \) and \( DCO \) are similar, and the product between the lengths respectively proportional starting from the point \(O\) is worth:
$$\overline{OA} \times \hspace{0.05em} \overline{OB} =\hspace{0.05em} \overline{OC} \times \overline{OD}$$
Case 3: External point of a circle with one tangent
If a point \( O\) is the intersection between a rope \( (AC )\) of a circle and a tangent to the same circle passing through \( T\), and such as the following figure:
So, the two triangles \( OAT \) and \( CTO \) are similar, and the product between the lengths respectively proportional starting from the point \(O\) is worth:
$$\overline{OA} \times \hspace{0.05em} \overline{OC} =\hspace{0.05em} \overline{OT} ^2$$
Case 4: External point of a circle with two tangents
If a point \( O\) is the intersection between two tangents to the circle passing respectively through \( T\) and \( T'\), and such as the following figure:
So, the triangle \( OTT' \) is an isosceles triangle, and in this case:
$$\overline{OT} = \hspace{0.05em} \overline{OT'}$$
Let us consider a circle and two ropes \(AB\) and \(CD\) such as the following figure:
Point \(O\) is the point of intersection between the lines \((AB)\) and \((CD)\) inside the circle.
Angles \( \widehat{BAC}\) and \( \widehat{BDC}\) intercept the same arc \( \overset{\frown}{BC}\) on the circle. They are therefore equal.
$$ \widehat{BAC} = \widehat{BDC} = \alpha $$
In the same way, the angles \( \widehat{ABD}\) and \( \widehat{ACD}\) intercept the same arc \( \overset{\frown}{AD}\) on the circle.
$$ \widehat{ABD} = \widehat{ACD} = \beta $$
The two triangles \( AOC \) and \( DOB \) having two angles two by two respectively equal, they are similar.
We then have the following proportions:
$$ \frac{OA}{OC} = \frac{OD}{OB} $$
And as a result,
$$\overline{OA} \times \hspace{0.05em} \overline{OB} =\hspace{0.05em} \overline{OC} \times \overline{OD}$$
Let us consider a circle and two ropes \(AB\) and \(CD\) such as the following figure:
Point \(O\) is the point of intersection between the lines \((AB)\) and \((CD)\) outside the circle, this intersection forms an angle vertex \(\theta\).
Angles \( \widehat{BAC}\) and \( \widehat{BDC}\) intercept the same arc \( \overset{\frown}{BC}\) on the circle. They are therefore equal.
$$ \widehat{BAC} = \widehat{BDC} = \alpha $$
Consequently, the angle measures \( \widehat{ABO}\) and \( \widehat{DCO}\) are equals, being the third angle of the two respective triangles:
$$ \widehat{ABO} = \widehat{DCO} = \beta = \pi - (\alpha + \theta) $$
The two triangles \( ABO \) and \( DCO \) having two angles two by two respectively equal, they are similar.
We then have the following proportions:
$$ \frac{OA}{OC} = \frac{OD}{OB} $$
And as a result,
$$\overline{OA} \times \hspace{0.05em} \overline{OB} =\hspace{0.05em} \overline{OC} \times \overline{OD}$$
By returning to our previous figure by making tend the points \(B\) and \(D\) towards a new point \(T\), so that the straight line \((OT)\) be a tangent to the circle.
By constructing this new point \(T\), we retained the equivalence between the two angles \(\alpha\).
The two triangles \( OAT \) an \( CTO \) having two angles two by two respectively equal, they are similar.
We then have the following proportions:
$$ \frac{OA}{OT} = \frac{OT}{OC} $$
And as a result,
$$\overline{OA} \times \hspace{0.05em} \overline{OC} =\hspace{0.05em} \overline{OT} ^2$$
By returning to our previous figure by making tend the points this time the points \(A\) and \(C\) towards a new point \(T'\), so that the straight line \((OT')\) be a tangent to the circle.
By constructing this new point \(T'\), we retained the equivalence between the two angles \(\alpha\).
The two similar triangles having merged during sliding, we then find ourselves with a single isosceles triangle \(OTT'\).
And as a result,
$$\overline{OT} = \hspace{0.05em} \overline{OT'}$$