AD, BE and CF are three medians of a triangle ABC. Let us prove that the centriod of ABC and DEF are the same point.

Given: A triangle ABC, AD BE and CF are medians, let us join DEF to make a triangle and label the points where medians of ΔABC intersects the sides of ΔDEF. Let O be the intersection of medians of ΔABC i.e. O is centroid of ΔABC

To Prove: O is centroid of ΔDEF

Proof:

As, CF and BE are medians to sides AB and AC respectively,

F is mid-point of AB and E is mid-point of AC

By mid-point theorem [The line segment connecting the midpoints of two sides of a triangle is parallel to the third side]

EF || BC

⇒ FH || BD and HE || DC

In ΔABD and ΔAFH

∠FAH = ∠BAD [Common]

∠AFH = ∠ABD [Alternate angles]

ΔABD ∼ ΔAFH [By AA similarity criterion]

……[1] [Similar triangles have equal ratio of Corresponding sides]

In ΔADC and ΔAHE

∠HAE = ∠DAC [Common]

∠AHE = ∠ADC [Alternate angles]

ΔADC ∼ ΔAHE [By AA similarity criterion]

……[2] [Similar triangles have equal ratio of Corresponding sides]

From [1] and [2], we have

Here, BD = DC ∵ AD is median

⇒ FH = HE

⇒ DH is a median from D to side FH

Similarly, we can prove that EI and FG are medians, and as intersection of FH, EI and FG is O.

⇒ O is centroid for ΔDEF

⇒ ΔDEF and ΔABC have same centroid.

Hence, Proved!