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Question: The mid-points of AB and DC of parallelogram ABCD are E and F respectively. AF and EC intersect the diagonal BD at points P and Q respectively. BP:PD = How much?

The problem involves a parallelogram, where two internal lines intersect one of the diagonals. To find the ratio of BP:PD, we needs to apply the principles of geometry. The mid-point theorem or mid-line theorem in triangles will be relevant here. According to the mid-point theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to and half the length of the third side. Since E and F are the midpoints of AB and CD in parallelogram ABCD, EF is parallel to both AC and BD (since these are diagonals of a parallelogram and diagonals of a parallelogram bisect each other), and also its half in length. Diagonals of a parallelogram divide it into two congruent triangles. In this case, it would be ΔABD and ΔCDA. Since P and Q are the points of intersection on the diagonal BD, it divides triangle ABD into two smaller triangles - ΔABP and ΔPDA and triangle ECF into two smaller triangles - ΔECP and ΔPFQ. Now, since EF || BD and EF is half of AC and BD, P and Q will also be midpoints of AC and BD respectively (due to converse of midpoint theorem). As a result, BP = PD (P is midway on BD because Q is midway on AC) which implies BP:PD = 1:1.

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