Question

ABCD is a trapezium in which AB = 2 DC and
AB || DC. If AC and BD intersect at 0, prove that area
(∆ AOB) = 4 area (∆COD). [Fig. 12.104]​

Answer 1

Answer:

Given: ABCD is a parallelogram, E and F are mid points of AB and DC and a line GH intersects AD, EF and BC at G, P, and H respectively.

Construction: Join GB and Let it intersect EF of X.

Now since E and F are mid points of AB and DC

⇒ AE = EB = AB

and DF= FC = DC

but AB = DC

⇒ AEFD and BCFE are parallelogram

⇒ AD EF BC ........ (1)

Now In ∆ ABG

AE = EB and EX AG (from (1))

⇒ X is the mid point of GB (mid point theorem)

and in ∆ GBH

GX = XB (∵ X is the mid point of GB)

and XP BH (from (1))

⇒ P is the mid point of GH

⇒ GP = PH

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