In △ ABC, D, E and F are mid-points of sides AB, BC and AC respectively. Prove that AE and DF bisect each other.

Given: ABC is a triangle and D, E and F are mid-points of sides AB, BC and AC respectively.

To prove: AE and DF bisect each other.

Construction: Join AE and DF.

Proof: Since D and F are the midpoints of AB and AC, by the midpoint theorem,

DF ∥ BC and DF = $\dfrac{1}{2}$ BC.

⇒ BC = 2DF

Now, since E is the midpoint of BC, we can write:

⇒ BE = EC ……………….(1)

Similarly, let G be the point of intersection of AE and DF. We need to show that G is the midpoint of both segments.

From the Midpoint Theorem applied in Δ ABE and Δ ACE:

BE = 2DG and EC = 2GF

⇒ 2DG = 2GF (from equation (1))

⇒ DG = GF

Thus, G is the mid-point of AE.

Also, since G lies on AE and AG = GE, G is also the midpoint of AE.

Hence, AE and DF bisect each other.