A triangle has side lengths: 3, 4, 5. Calculate the radius of the inscribed circle (the circle inside the triangle and tangent to all three sides of the triangle).
Let's call $A, B, C$ the vertices of the triangle, $a=BC, b=CA, c=AB$ the lengths of the three sides, $I$ and $r$ the center and the radius of the inscribed circle. The heights from $I$ to the triangles $IAB, IBC$, and $ICA$ are all equal to $r$, so the areas of these three triangles are: area $(IAB)=\frac{rc}{2}$, area $(IBC)=\frac{ra}{2}$, area $(ICA)=\frac{rb}{2}$, from which we deduce: $\operatorname{area}(ABC)=\frac{r(a+b+c)}{2}$. The triangle with sides $3,4,5$ is a right triangle, its area is: $\frac{3 \times 4}{2}=6=\frac{r(3+4+5)}{2}$. It follows that $r=1$.

The given triangle, with sides $3,4,5$, being a right triangle, we can also reason as follows. Let $D, E, F$ be the projections of $I$ onto the sides $BC, CA, AB$. These are the points of contact of the inscribed circle with the sides of the triangle. If $A$ is the vertex of the right angle, $AEIF$ is a square, and $r=IE=IF=AE=AF$. Now, $AE$ and $AF$ are easy to calculate: if we set $u=AE=AF, v=BF=BD$, $w=CD=CE, u+v=c, v+w=a, w+u=b$, then by adding: $2(u+v+w)=a+b+c$ or finally $u=\frac{-a+b+c}{2}, v=\frac{a-b+c}{2}, w=\frac{a+b-c}{2}$. We deduce: $r=u=\frac{-5+4+3}{2}=1$.

The two methods are very different, but the results are the same. Indeed, $\frac{r(a+b+c)}{2}=\left(\frac{-a+b+c}{2}\right)\left(\frac{a+b+c}{2}\right)=\frac{-a^{2}+(b+c)^{2}}{4}=\frac{bc}{2}=$ area $(ABC)$ because the square of the hypotenuse $a^{2}=b^{2}+c^{2}$.

Let's call $M$ and $N$ the midpoints of $[AB]$ and $[CD]$. Suppose first that $(AD)$ and $(BC)$ intersect at a point $S$, and consider the line $(SM)$, which intersects $(CD)$ at $P$. Using Thales' theorem and the fact that $MA=MB$, we have: $\frac{PD}{PS}=\frac{MA}{MS}=\frac{MB}{MS}=\frac{PC}{PS}$, which implies: $PC=PD$. Therefore, $P$ is the midpoint $N$ of $[CD]$, and the line $(SM)$ is none other than $\Delta: \Delta, (AD)$ and $(BC)$ are indeed concurrent at $S$.

Now suppose that $(AD)$ and $(BC)$ are parallel, and draw the line parallel to $(AD)$ and $(BC)$ passing through $M$. Let $P$ be the intersection of this parallel with $(CD)$. $DPM A$ and $CPMB$ are parallelograms, so $DP=MA=MB=CP$. Once again, $P$ is the midpoint of $[CD]$, so $(MP)$, parallel to $(AD)$ and $(BC)$, is none other than $(MN)=\Delta$.