In the DABC tetrahedron, the M-point of intersection of the medians of the BDC face

In the DABC tetrahedron, the M-point of intersection of the medians of the BDC face, E of the AC seridine. Expand the vector EM in vectors AC, AB, AD.

We construct a vector ↑ EK.

↑ ЕМ = ↑ ЕК + ↑ КМ.

EK is the middle line of the triangle ABC and is equal to half AB, KM = 1/3 DK by the property of medians, then:

↑ EM = ↑ AB / 2 – ↑ DK / 3.

↑ DC = (↑ DB + ↑ DC) / 2.

↑ EM = ↑ AB / 2 – ↑ DB / 6 – ↑ DS / 6.

↑ DB = ↑ DA + ↑ AB.

↑ DC = ↑ DA + ↑ AC.

Then: ↑ EM = ↑ AB / 2 – ↑ DA / 6 – ↑ AB / 6 – ↑ DA / 6 – ↑ AC / 6 = ↑ AB / 3 – ↑ DA / 3 – ↑ AC / 6 =

↑ AB / 3 + ↑ AD / 3 – ↑ AC / 6.

Answer: ↑ EM = ↑ AB / 3 + ↑ AD / 3 – ↑ AC / 6.