Centroid

Theorem: All the three medians of a triangle intersect at a single point called centroid, which divides each median in a ratio of $2:1$.

Prerequisites:
Similarity of triangles (definition)
Midpoint Theorem (proof)
AA similarity (proof)
Alternate angles property (proof)
vertical angle theorem (proof)

Proof:

Let ABC be a triangle having medians AD, BE and CF.
Let us consider the medians BE and CF, which intersect at a point G. Let us join E and F by a straight line.

Since EF joins the midpoints of the lines AB and AC, hence by midpoint theorem:

$\qquad\quad EF \parallel BC$
And, $\;\;\;\! EF = \dfrac{1}{2} BC\qquad\qquad\qquad\cdots\text{(1)}$

Now, consider triangles $\triangle BCG$ and $\triangle EFG$:

$\qquad\quad\angle EGF = \angle BGC\qquad\qquad\qquad\text{(vertically opposite angles)}\\ \qquad\quad\angle GFE = \angle GCB\qquad\qquad\qquad\text{(alternate angles)}\\ \therefore\quad\;\;\triangle BCG\sim\triangle EFG\qquad\qquad\qquad\text{(AA similarity)}$

Thus, by definition of similarity, corresponding sides are proportional, hence:

$\qquad\quad\dfrac{GE}{GB} = \dfrac{GF}{GC} =\dfrac{EF}{BC} = \dfrac{1}{2}\qquad\qquad\text{(from $(1)$)}$

Thus, G divides BE and CF in the ratio $2:1$.

Similarly, by considering the medians AD and BE, which intersect at a point G', it can be shown that G' divides AD and BE in the ratio 2:1.

But BE is divided in $2:1$ ratio by G. Hence, G' $=$ G.

Thus, all the medians intersect at a single point, which divides the medians in a ratio $2:1$.

Q.E.D.


Recommended;
Median through hypotenuse
Basic Proportionality Theorem
Angle Bisector Theorem