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ABCD is a parallelogram .E is the mid point of BC.DE meets diagonal AC at Oand meets AB(produced) at F.prove that

(1) DO:OE=2:1
(2)area of triangle OEC:area of triangle OAD=1:4
please help me with this question



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CE=1/2 BC      (As E is mid point of BC)
CE=1/2 AD      (AD=BC)
2CE=AD          ...1
angle AOD = angle COE    (Vertically opposite angles)
angle OAD = angle OCE    (alternate interior angles)
So, ΔOAD similar to ΔOCE by AA Similarity
So by c.p.c.t. , OA/OC=AD/CE=OD/OE
So, OD:OE=2:1
As ΔOAD similar to ΔOCE, So by area theorem for similar triangle
area of ΔOEC/area of ΔOAD = (CE/AD)²= (CE/2CE)²= (1/2)²
Area of ΔOEC/Area of ΔOAD = 1/4
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