Question

Diagonals AC and BD of quadrilateral ABCD intersect at 'O', such that OB = OD. If AB = CD, then show that.

(i) Ar $\left(\Delta DOC\right)$ = AR $\left(\Delta AOB\right)$

(ii) Ar $\left(\Delta DCB\right)$ = Ar $\left(\Delta ACB\right)$

(iii) AD || BC

Answer 1

(i) In triangles DOE and BOF, we have

OB = OD

$\angle$ DOE = $\angle$ BOF

$\angle$ DEO = $\angle$ BFO

So, by AAS congruence criterion, we have

$△$ DOE $\cong$ $△$ BOF

$⟹ar\left(△DOE\right)=ar\left(△BOF\right)$ and $DE=BF$ ...(1)

In triangles DEC and BFA, we have

$\angle$ DEC = $\angle$ BFA = 90${}^o$

DE=BF

CD=AB

So, by RHS congruence criterion, we have

$△$ CDE $\cong$ $△$ ABF

$⟹ar\left(△CDE\right)=ar\left(△ABF\right)$ and $\angle DCE=\angle BAF$ ....(2)

From 1 & 2, we have

$ar\left(△DOE\right)+ar\left(△CDE\right)=ar\left(△BOF\right)+ar\left(△ABF\right)$

$ar\left(△COD\right)=ar\left(△AOB\right)$

(ii) We have,

$ar\left(△AOB\right)=ar\left(△COD\right)$

$ar\left(△AOB\right)+ar\left(△BOC\right)=ar\left(△COD\right)+ar\left(△BOC\right)$

$ar\left(△ACB\right)=ar\left(△DCB\right)$

(iii) From 2, we have

$\angle DCE=\angle BAF$

$\angle ACD=\angle CAB$

$DC\parallel AB$

Also, DC=AB

So, ABCD is a parallelogram.

Hence, $DA\parallel CB$.