Question

State true or false:

In the given figure, the diagonals $AC$ and $BD$ intersect at point $O$ . If $OB=OD$ and $AB//DC$, then

$Area(△DCB)=Area(△ACB)$ .

• A. True
• B. False

Answer 1

The correct option is A True

Given, the diagonals AC and BD intersect at point O such that OB$=$OD and AB$\parallel$DC.
Now, In $△DOC$ and $△AOB$
$\angle BDC=\angle DBA$ (Alternate angles, As AB$\parallel$CD and BD intersects it.)
$OD=OB$ (Given)
$\angle DOC=\angle AOB$ ( Vertically opposite angles)
So, $△DOC\cong △AOB$ (By ASA concurrence Property)
Sides of concurrent triangles are equal.
So, AB $=$ DC. --(1)
As, AB$\parallel$ DC and AB$=$ ,ABCD becomes a parallelogram.(A quadrilateral having one side equal and parallel is a parallelogram).
Now, In $△ACB$ and $△DCB$
$AB=CD$ From (1)
$BCiscommon$
$AC=BD$ (As diagonals of parallelogram are equal)
So, $△ACB\cong △DCB$ (By SSS concurrence Property)
Since, Concurrent triangle have equal areas.
$\therefore$ Area of $△ACB=$ Area of $△DCB$