Question

In a given figure 5.5 $\square$LMNO is a quadrilateral in which $\angle$LMN = $\angle$MLO = 90$°$ seg LO $\cong$ seg MN then prove that
(i) $\angle$MNO $\cong$ $\angle$LON
(ii) seg ON $\cong$ seg LM
(iii) $\angle$ MNO = 90$°$

Answer 1

Consider quadrilateral LMNO,
$\angle$MLO + $\angle$NML = 90$°$ + 90$°$
= 180$°$
Since, angles on the same side of a transversal line are supplementary,
$\therefore$seg LO$\parallel$seg MN
We know that if a pair of opposite sides in a quadrilateral are equal and parallel, then it is a parallelogram.
Hence, LMNO is a parallelogram.


$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{As},\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{opposite}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{parallelogram}\mathrm{are}\mathrm{equal}. \\ \therefore \angle \mathrm{NML}=\angle \mathrm{LON}=90° \\ \&\angle \mathrm{ONM}=\angle \mathrm{MLO}=90° \end{gathered}$$

$\therefore$$\angle$MNO $\cong$ $\angle$LON
(ii) seg ON $\cong$ seg LM (Opposite sides of a parallelogram are equal.)
(iii) $\angle$MNO = $\angle$MLO = 90$°$ (Opposite angles of a parallelogram are equal.)