PQRS is a square. N and M are midpoints of SR and QR respectively. O is the midpoint of diagonal PR. show that NOMR is a square.

Given   PQRS is a square. N and M are mid points of SR and QR respectively .O is the mid point of PR.
Let the length of the sides of square be a units.

$$\begin{gathered} So PQ=QR=RS=SP = a units \\ Now length of the diagonal PR = \sqrt{P{Q}^2+Q{R}^2}=\sqrt{a^2+a^2}=a\sqrt{2} units \\ And as O is the midpoint of PR \\ \therefore OR =\frac{a\sqrt{2}}{2} =\frac{a}{\sqrt{2}}units \\ Now as N and M are the midpoints . \\ Hence MR = RN = \frac{a}{2}units \\ Now from midpoint theorem we can say that line joining the midpoints of two sides of a triangle \\ is parallel to the third side and equals half the length of third side \\ Hence in ∆RPQ, OM\parallel PQ and in∆RPS,ON\parallel SP \\ \therefore \angle RMQ =\angle RQP=90° and \angle RNO=\angle RSP = 90° \\ Now in ∆ORM ,we have MR = a units and OR = \frac{a}{\sqrt{2}}units \\ So OM =\sqrt{O{R}^2-M{R}^2} =\sqrt{(\frac{a}{\sqrt{2}}{)}^2-(\frac{a}{2}}{)}^2 =\frac{a}{2} units \\ Similarly we can find ON = \frac{a}{2}units \\ Now we have the quadrilatera\mathrm{l} \mathrm{NOMR} \mathrm{having} \mathrm{all} \mathrm{sides} \mathrm{equal} \mathrm{and} \mathrm{measures} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{angles} \mathrm{is} 90° \\ \therefore \mathrm{NOMR} \mathrm{is} \mathrm{a} \mathrm{square} \end{gathered}$$