# In tetrahedron ABCD point M, K, P midpoints of edges AB, BD, BC prove that plane MKP

**In tetrahedron ABCD point M, K, P midpoints of edges AB, BD, BC prove that plane MKP is parallel to plane ACD and Find the area of triangle MKP if the area of triangle ACD is 48cm².**

All edges of a tetrahedron are equal. Since, according to the condition, points M, K, P are the middle of the segments AB, BD, BC, then the segment KM is the middle line of the triangle ABD, KP is the middle line of the triangle BCD, MP is the middle line of the triangle ABC.

The midline segments are parallel to the base of the triangles: MK || АD, КР || СD, МР || AC, then the MCR plane is parallel to the ACD plane, which was required to be proved.

The length of the middle line of the triangle is equal to half the length of the parallel side, then the MCP triangle is similar to the ACD triangle in three proportional sides with the similarity coefficient K = AD / MK = AD / (AD / 2) = 2.

The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.

Savs / Smcr = 48 / Smcr = 22.

Smcr = 48/4 = 12 cm2.

Answer: The area of the MCR triangle is 12 cm2.