Answer to Question #90266 in Trigonometry for Aravind

Question

If a triangle is isosceles then the two median towards that equal sides are of equal length. Conversely if 2 medians of a triangle are equal in length, the triangle is isosceles. Prove.

Answer 1

I. ∆ ABC,

AC = BC,

AK and BF are medians.

Consider the triangles ∆ ACK and ∆ BCF.

1) AC = BC (by condition as the sides of an isosceles triangle))

2) CK = CF (since the medians AK and DF are drawn to equal sides AC and BC, then half of these sides are equal to each other)

3) ∠C is common.

Therefore, ∆ACK = ∆BCF (on two sides and the angle between them).

The equality of the triangles implies the equality of the corresponding sides: AK = BF.

II.In the triangle ∆ ABC,

AK and BF are medians,

AK=BF,

O is the intersection point of the medians.

It is known that the medians are divided by the intersection point in the ratio 2: 1. It means:

OF = BF / 3 =AK / 3 = OK,

AO = BO.

Consider the triangles ∆ AOF and ∆ KOB:

1) AO = OB,

2) OF = OK,

3)∠ AOF = ∠ KOB (vertical angles).

Thus, the triangles ∆ AOF and ∆ KOB are equal, then AF = KB =AC / 2 = BC / 2, hence

AC = BC.