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Question

In the given figure, ABCD is a parallelogram. P is the mid-point of AB. Prove that ACBM is a parallelogram.
1112148_9d26ac8a41984f9bb7511ac5e72c41d4.png

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Solution

(i) ABCD is parallellogram
(ii) P is midpoint pf CM.
ABCD is parfallellogram
AD||BC .
MAB=CBA ,,,(1) (interior opposite angle of tranversal AB)
CPB=ABM ...(2) (vertical opposite angle)
In ΔCPB and ΔABM
from (1) and (2) BCP=PMA
CP+PM=CM2 ...(4) (P is mid point )
from (2), (3) and (4)
ΔCPBΔMPA
CB=AM
We know CB is parallel to AM
AMBis a parallelllogram sinceCB||AMandCB=AM

1089370_1112148_ans_bd789725330e4c95ade5fbc16afe247e.png

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