Let $\mathcal{C}$ be a circle, and $A, B, C, D$ four points on it. Suppose that $(AB)$ and $(CD)$ intersect at a point $M$. Find two similar triangles.
We can assume that $A, B, M$ and $C, D, M$ are collinear in this order. In this case, $M A B$ and $M D C$ are similar.

Note: We thus have $M A \times M B = M C \times M D$. If the point $M$ is fixed, for any line passing through $M$ and intersecting the circle at $E$ and $F$, the product $M E \times M F$ is therefore constant. This quantity is called the power of $M$ with respect to the circle $\mathcal{C}$. A course on this subject can be found in the old geometry polycopied.

Solution to exercise $3$: $B, C^{\prime}, B^{\prime}, C$ are on the circle with diameter $[B C]$, so $\widehat{A C B} = \widehat{B^{\prime} C B} = 180^{\circ} - \widehat{B^{\prime} C^{\prime} B} = \widehat{A C^{\prime} B^{\prime}}$. Since $\widehat{C^{\prime} A B^{\prime}} = \widehat{A B C}$, we deduce that $A B C$ and $A C^{\prime} B^{\prime}$ are similar. Similarly, $B A^{\prime} C^{\prime}$ and $B C A$ are similar, and $C A^{\prime} B^{\prime}$ and $C B A$ are also similar.

Moreover, $\widehat{H C A^{\prime}} = 90^{\circ} - \widehat{C^{\prime} B A^{\prime}} = \overline{C^{\prime} A H}$, so the right triangles $B A A^{\prime}$ and $H C A^{\prime}$ are similar.

Solution to exercise $4$: $\frac{M E}{M F} = \frac{D A}{D C}$ if and only if $M E F$ and $D A C$ are similar. We note that $A, E, F, M$ are on the circle with diameter $[A M]$, so $\widehat{M E F} = \widehat{M A F} = \widehat{D A C}$ by the inscribed angle theorem. Similarly, $\widehat{E F M} = \widehat{E A M} = \widehat{B A C}$. But $\widehat{B A C} = \widehat{A C D}$ because they are alternate interior angles. Thus $M E F$ and $D A C$ are similar, which concludes.

Solution to exercise $5$: $A, D, F, M$ are concyclic, because $\widehat{A F M} = \widehat{A D M} = 90^{\circ}$. Thus $\widehat{M D F} = \widehat{M A F}$. Similarly, $\widehat{M E D} = \widehat{M B D} = \widehat{M B A}$ because $B, D, E, M$ are concyclic. Finally, $\widehat{M A F} = \widehat{M B A}$ because $(A F)$ is tangent to the circumcircle of $M B A$. Thus $\widehat{M D F} = \widehat{M E D}$.

Similarly, we prove that $\widehat{M D E} = \widehat{M F D}$. Therefore, $M D F$ and $M E D$ are similar, which implies that $M D / M F = M E / M D$, or $M D^2 = M E \times M F$.