The diagonals of the parallelogram ABCD meet at point O. Point P is such that DOCP is also a parallelogram (CD is its diagonal). Let Q denote the intersection point of BP and AC, and R – the intersection point of DQ and CP. Prove that PC = CR.
The solution of the problem ;
Note that the segments DP and BC are parallel and equal;
Therefore BOPC is a parallelogram, whence QC = OC / 2 = PD / 2;
Thus, the segment QC with ends on the sides RD and RP of the triangle DRP is parallel to the side DP of this triangle and is equal to its half;
Hence, it is the middle line of this triangle;
Consequently, C is the midpoint of RP, as required.
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