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In Euclid's geometry context, I have the following problem: Let ABC a triangle and P laying in AB. We need to find a point Q in AC or BC such that the triangle APQ has the half of the area of ABC. Let M be the medium point of AB. Construct a parallel segment to PC passing by M, and call Q it's intersection with AC (or BC). Prove that Q is the point we find.

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The median $CM$ divides $\triangle ABC$ into two triangles of equal area. If $P=M$ we are finished. We can suppose without loss of generality that $P$ lies between $A$ and $M$.

Draw a picture.

Note that the area of $\triangle CMB$ is the area of $\triangle MBQ$ plus the area of triangle $MQC$.

But the area of $\triangle MQC$ is the same as the area of $\triangle MQP$, since they have the same base $MQ$, and by parallelism the same height.

Thus the area of $\triangle PBQ$ is the same as the area of $\triangle CMB$, that is, half the area of $\triangle ABC$. This completes the proof.

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