  Question

In a parallelogram $$ABCD,\ E$$ and $$F$$ are the mid-points of sides $$AB$$ and $$CD$$ respectively (see Fig). Show that the line segments $$AF$$ and $$EC$$ trisect the diagonal $$BD$$. Solution

$$ABCD$$ is parallelogramUsing properties of a parallelogram $$AB \parallel DC$$ and $$AB = DC$$$$E$$ is the mid-point of $$AB$$$$AE=\dfrac{1}{2} AB$$    ... (1)$$F$$ is the mid-point of $$CD$$Therefore, $$CF=\dfrac{1}{2 }CD$$$$CF=\dfrac{1}{2} AB$$ (Since $$CD = AB$$)    ... (2)From (1) and (2), we get$$AE = CF$$Also,$$AE \parallel CF$$ (Since $$AB \parallel DC$$)Thus, a pair of opposite sides of a quadrilateral  $$AECF$$ are parallel and equal.Quadrilateral $$AECF$$ is a parallelogram.$$EC \parallel AF$$$$EQ \parallel AP$$ and $$QC \parallel PF$$In $$\triangle BPA$$,  $$E$$ is the mid-point  of $$BA$$         $$EQ \parallel AP$$ $$\mid$$ Using mid point theorem $$BQ = PQ$$          ... (3)Similarly, by taking  $$\triangle CQD$$,  we can prove that$$DP = QP$$    ... (4)From (3) and (4), we get$$BQ = QP = PD$$Therefore,$$AF$$ and $$CE$$ trisect the diagonal $$BD$$.MathematicsRS AgarwalStandard IX

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