5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.
Solution: Given that, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. To show, AF and EC trisect the diagonal BD. Proof, ABCD is a parallelogram , AB || CD also, AE || FC Now, AB = CD (Opposite sides of parallelogram ABCD) ⇒½ AB = ½ CD ⇒ AE = FC (E and F are midpoints of side AB and CD) AECF is a parallelogram (AE and CF are parallel and equal to each other) AF || EC (Opposite sides of a parallelogram) Now, In ΔDQC, F is mid point of side DC and FP || CQ (as AF || EC). P is the mid-point of DQ (Converse of mid-point theorem) ⇒ DP = PQ — (i) Similarly, In ΔAPB, E is midpoint of side AB and EQ || AP (as AF || EC). Q is the mid-point of PB (Converse of mid-point theorem) ⇒ PQ = QB — (ii) From equations (i) and (i), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Hence Proved.
