# A B points in plane. M symetric of A face of B , N symetric of B face of A. show MN and AB have same midpoint.

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The problem provides the information that M is symmetric of A with respect to B, hence, B is the midpoint of line segment AM.

Using the position vectors `bar r_A, bar r_B, bar r_M` , yields:

`bar r_B = (1/2)(bar r_A + bar r_M) ` `=> bar r_M = 2bar r_B - bar r_A`

The problem provides the information that N is symmetric of B with respect to A, hence, A is the midpoint of line segment BN.

`bar r_A = (1/2)(bar r_B + bar r_N) => bar r_N = 2bar r_A - bar r_B`

Supposing that P is the midpoint of AB yields:

`bar r_P = (1/2)(bar r_A + bar r_B)`

Supposing that P' is the midpoint of MN yields:

`bar r_P' = (1/2)(bar r_M + bar r_N)`

Replacing `2bar r_B - bar r_A` for `bar r_M` and `2bar r_A - bar r_B` for `bar r_N` , yields:

`bar r_P' = (1/2)(2bar r_B - bar r_A + 2bar r_A - bar r_B)`

`bar r_P' = (1/2)(bar r_A + bar r_B) = bar r_P`

**Hence, the midpoint P coincides with the midpoint P', hence, the line segments MN and AB share the midpoint P.**