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!

#### Answers

* HERE,IN TRIANGLES AOD AND BOC, *

* OB=OD *

* AB=CD *

* ANGLE AOD= ANGLE BOC[V.O.A.] *

* HENSE,AOD CONGRUENT TO BOC *

* OA=OC[C.P.C.T.] *

* 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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