Diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that ABCD is a llgm.. It 's urgent ASAP!
HERE,IN TRIANGLES AOD AND BOC,
ANGLE AOD= ANGLE BOC[V.O.A.]
HENSE,AOD CONGRUENT TO BOC
HENCE,ABCD IS A QUAD. IN WHICH DIAGONAL AC AND BD BISECT EACH OTHER
SO,ITS A llgm.
hope u got it!
i thought of that already..... but its an S.S.A. and that 's not allowed
In the quad., the diagonals intersect at O such that OB=OD.In triangle BDC O is the mid-point of BD.
Construction: A line from O meeting BC at X and parallel to DC.
Due to the converse of mid-point theorem,X is the mid-point of BC.
In triangle CAB,O is the mid-pt. of AC and X is the mid-pt. of BC.Therefore due to mid-point theorem OX//AB.
OX//AB and OX//DC.Thus AB//CD.
AB=CD and AB//CD.Thus ABCD is a //gm(Proved)
Given, a quadrilateral ABCD with AB = DC and DO = OB.
To prove: ABCD a parallelogram
Construction: Draw OX DC.
In Δ BCD,
O is the mid point of BD and OX DC.
∴By converse of mid point theorem,
X is the mid point of BC and OX = DC.
As AB = DC
∴ OX = AB.
Now, in Δ ABC,
X is the mid point of BC and OX = AB.
∴ By converse of mid point theorem,
O is the mid point of AC and OX AB
As OX AB and OX DC
⇒ AB DC
Now, AB DC and AB = DC
∴ ABCD is a parallelogram.
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