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I know that a median of a triangle is a line joining one
of the vertices to the mid-point of the opposite side.

For example, in a triangle OAB, O is the origin, $A$ is the point $(0,6)$ and $B$ is the point $(6,0)$.

How can I show that the point $(2,2)$ lies on all three medians?

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$\color{red}{\text{Method 1}}$: All three medians of a triangle pass through the centroid of the triangle hence the centroid of $\triangle OAB$ having vertices $O(0, 0)$, $A(0, 6)$ & $B(6, 0)$ is given as $$\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\equiv\left(\frac{0+0+6}{3}, \frac{0+6+0}{3}\right)\equiv(2, 2)$$ Thus the given point $(2, 2)$ (coincident with the centroid) lies on all three medians of $\triangle OAB$.

$\color{red}{\text{Method 2}}$: Calculating the mid-points of all the sides of $\triangle OAB$ & finding the equations of all three medians of $\triangle OAB$ as follows

$\color{blue}{\text{Median 1}}$: drawn from the vertex $A(0, 6)$ to the mid-point of side $OB$ i.e. $\left(\frac{0+6}{2}, \frac{0+0}{2}\right) $ or $(3, 0) $ given as $$y-6=\frac{0-6}{3-0}(x-0)$$ $$\implies \color{blue}{2x+y-6=0}$$
Substituting the coordinates of the given point $(2, 2)$ in the above equation of the median, as follows $$2(2)+2-6=0$$ $$\implies 0=0$$ Hence, point $(2, 2)$ lies on the median through the vertex $A$. Similarly,

$\color{blue}{\text{Median 2}}$: drawn from the vertex $B(6, 0)$ to the mid-point of side $OA$ i.e. $\left(\frac{0+0}{2}, \frac{0+6}{2}\right) $ or $(0, 3) $ given as $$y-0=\frac{3-0}{0-6}(x-6)$$ $$\implies \color{blue}{x+2y-6=0}$$
Substituting the coordinates of the given point $(2, 2)$ in the above equation of the meridian, as follows $$2+2(2)-6=0$$ $$\implies 0=0$$ Hence, point $(2, 2)$ lies on the median through the vertex $B$.

$\color{blue}{\text{Median 3}}$: drawn from the vertex $O(0, 0)$ to the mid-point of side $AB$ i.e. $\left(\frac{0+6}{2}, \frac{6+0}{2}\right) $ or $(3, 3) $ given as $$y-0=\frac{3-0}{3-0}(x-0)$$ $$\implies \color{blue}{y=x}$$
Substituting the coordinates of the given point $(2, 2)$ in the above equation of the median, as follows $$2=2$$ Hence, point $(2, 2)$ lies on the median through the vertex $O$.

Thus the given point $(2, 2)$ lies on all three medians of $\triangle OAB$

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The midpoint between $A$ and $B$ is $D(3,3)$, so an equation of the line $DO$ is $y=x$. Here, $(2,2)$ is on this line.

The midpoint between $B$ and $O$ is $E(3,0)$, so an equation of the line $EA$ is $y-0=\frac{0-6}{3-0}(x-3)$, i.e. $y=-2x+6$. Here, $(2,2)$ is on this line.

The midpoint between $A$ and $O$ is $F(0,3)$, so an equation of the line $FB$ is $y-3=\frac{3-0}{0-6}(x-0)$, i.e. $y=-\frac 12x+3$. Here, $(2,2)$ is on this line.

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Centroid is a point which lies on all three medians, and its coordinates are $$ \frac13 (x_O + x_A + x_B,\, y_O + y_A + y_B) = (2, 2) $$

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The equation of the straight line joining $O$ with the middle of the segment [$(0,6)$,$(6,0)$] ($(3,3)$) is

$$y=x.$$

The equation of the straight line joining the middle of the segment [$O$,$(0,6)$], ($(0,3)$ ) with $(6,0)$ is

$$y=-\frac{1}2x+3$$.

See the following figure:

enter image description here

Equating the two equations, we get $x=y=2$.

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