Medians AM and BN in triangle ABC meet at point O. Prove that triangle AOB is similar to triangle MON.

The medians at the intersection divide each other in a 2: 1 ratio:

BO = 2 * ON;

AO = 2 * OM;

Segment MN – the middle line for ΔABC (connects the midpoints of the AC and BC sides). If MN is the middle line, then MN is half of the side AB:

MN = AB / 2;

AB = 2 * MN.

Each side of ΔABC is twice as large as the corresponding sides of ΔMON, that is, the sides of ΔABC are proportional to the sides of ΔMON with a proportional factor of 2. Triangles with proportional sides are similar.

Method 2.

The center line NM is parallel to side AB.

<OAB = <AMN; <ABO = <BNM;
<AOB = <NOM (vertical).

The 3 corners of one triangle are equal to the 3 corners of the other triangle.