Calculate the medians of a triangle with sides 13 cm, 13 cm, 10 cm

Given: △ ABC, AB = BC = 13 cm, AC = 10 cm.
Find: medians BE, СF, AK.
Since the sides AB and BC of this triangle are equal, then △ ABC is isosceles.
And this means that the median BE, drawn from the vertex B to the base of the triangle AC, is also its bisector and height. Hence ∠Е = 90 ° and, accordingly, the formed △ ABE is rectangular.
Find the AE leg:
AE = 1/2 AC = 10/2 = 5 (cm).
Behind the Pythagorean theorem:
AB ^ 2 = AE ^ 2 + BE ^ 2.
Hence:
BE ^ 2 = AB ^ 2 – AE ^ 2 = 169 – 25 = 144 (cm).
BE = Sqrt144 = 12 (cm).
By the property of the medians:
OE = 1/3 BE = 12/3 = 4 (cm).
In △ AOE, following the Pythagorean theorem:
AO ^ 2 = AE ^ 2 + OE ^ 2 = 25 + 16 = 41 (cm).
AO = Sqrt41 ≈ 6.4 (cm).
By the property of the medians, AO = 2/3 AK. Hence:
AK = AO * 3/2 = 6.4 * 3/2 = 9.6 (cm).
Since △ ABC is isosceles, then CF = AK = 9.6 (cm).
Answer: BE = 12 cm, CF = 9.6 cm, AK = 9.6 cm.