Consider an equilateral △ABC, and Let D, E, F are midpoints of BC, CA and AB.
Here, AD, BE and CF are medians of △ABC.
Now,
D is midpoint of BC => BD = DC
Similarly, CE = EA and AF = FB
Since ΔABC is an equilateral triangle
AB = BC = CA …..(i)
BD = DC = CE = EA = AF = FB …………(ii)
And also, ∠ ABC = ∠ BCA = ∠ CAB = 60° ……….(iii)
Consider Δ ABD and Δ BCE
AB = BC [From (i)]
BD = CE [From (ii)]
∠ ABD = ∠ BCE [From (iii)]
By SAS congruence criterion,
Δ ABD ≃ Δ BCE
=> AD = BE ……..(iv) [Corresponding parts of congruent triangles are equal in measure]
Now, consider Δ BCE and Δ CAF,
BC = CA [From (i)]
∠ BCE = ∠ CAF [From (ii)]
CE = AF [From (ii)]
By SAS congruence criterion,
Δ BCE ≃ Δ CAF
BE = CF …………..(v) [Corresponding parts of congruent triangles are equal]
From (iv) and (v), we have
AD = BE = CF
Median AD = Median BE = Median CF
The medians of an equilateral triangle are equal.
Hence proved.