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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 $$BD$$.

463893_a2f380c87d2d4fbda500a37f15f79298.png


Solution

$$ABCD$$ is parallelogram

Using 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 $$BD$$.


Mathematics
RS Agarwal
Standard IX

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