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Question
If L and M are the mid-points of AB, and DC respectively of parallelogram ABCD. Prove that segment DL and BM trisect diagonal AC.
Solution
Since L and M are the mid-points of AB and Dc respectively.
`"BL" = (1)/(2)"AB" and "Dm" = (1)/(2)"DC"`....(i)
But ABCD is a parallelogram
Therefore, AB = CD and AB || DC
⇒ BL = DM and BL || Dm ...(from (i))
⇒ BLDM is a parallelogram.
⇒ DL || Dm
⇒ LP || BQ ............(ii)
It is known that the segment drawn through the mid-point of one side of a triangle and parallel to the other side bisects the third side.
In ΔABQ , L is the mid-point of AB and MQ || PD
Therefore, P is mid-point of AQ
Hence, AP = PQ ..........(iii)
Similarly, in ΔCPD, M is the mid-point of CD and LP || BQ
Therefore, Q is mid-point of CP
Hence, PQ = QC ..........(iv)
From (iii) and (iv)
AP = PQ = QC
Therefore, P and Q trisect AC
Thus, DL and BM trisect AC.
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