# Q: 6     In Fig. $\small 9.25$, diagonals AC and BD of quadrilateral ABCD intersect at O such that. If $\small AB=CD$, then show that:             (i)  $\small ar(DOC)=ar(AOB)$            [Hint: From D and B, draw perpendiculars to AC.]

M manish

We have ABCD is quadrilateral whose diagonals AC and BD intersect at O. And OB = OD, AB = CD
Draw DE $\perp$ AC and FB $\perp$AC

In $\Delta$DEO and $\Delta$ BFO
$\angle$DOE = $\angle$BOF [vertically opposite angle]
$\angle$OED = $\angle$BFO [each $90^0$]
OB = OD [given]

Therefore, by AAS congruency
$\Delta$DEO $\cong$  $\Delta$ BFO
$\Rightarrow$ DE = FB [by CPCT]

and ar( $\Delta$DEO) =  ar($\Delta$BFO) ............(i)

Now, In  $\Delta$DEC and  $\Delta$ABF
$\angle$DEC = $\angle$BFA [ each $90^0$]
DE = FB
DC = BA [given]
So, by RHS congruency
$\Delta$DEC $\cong$  $\Delta$ BFA
$\angle$1 = $\angle$2 [by CPCT]
and, ar( $\Delta$DEC) =  ar($\Delta$BFA).....(ii)

By adding equation(i) and (ii), we get
$\small ar(DOC)=ar(AOB)$
Hence proved.

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