In an arbitrary pentagon ABCDE, points: K – middle AB, L – middle BC, M – middle CD, N – middle DE.

In an arbitrary pentagon ABCDE, points: K – middle AB, L – middle BC, M – middle CD, N – middle DE. Points: Р – middle of KM, Q – middle of LN. Prove: a) PQ || AE; b) PQ = 1/4 AE

Connect the vertices A and D of the pentagon, and mark R on the segment AD – the middle of AD.

Let’s connect the points L with R. Then in the quadrilateral ABCD the segments LR and KM are midline of the quadrangle, the point of intersection of which is their common midpoint.

Then LP = PR, and since Q is the midpoint of LN, then PQ is the midline of triangle LRN.

PQ = RN / 2. PQ || RN as the centerline and base of the triangle.

In triangle ADE, the segment RN is its middle line, then RN = AE / 2, and since RN = 2 * PQ, then PQ = AE / 4.

RN || AE, and since PQ || RN, then PQ || AE, as required.