In figure it is given that AB and CD are two parallel lines intersected by a transversal EF.Bisectors of interior angles <BMN and <DNM on the same side of the transversal meet at P. Prove that <MNP=90 degrees.​

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$$\begin{gathered} \mathrm{Some} \mathrm{correction} \mathrm{that} \mathrm{we} \mathrm{have} \mathrm{to} \mathrm{prove} \mathrm{that} \angle \mathrm{MPN}=90° \\ \Rightarrow \angle \mathrm{EMB}+\angle \mathrm{BMN}=180° \left(\mathrm{angle} \mathrm{of} \mathrm{straight} \mathrm{is} 180°\right) \\ \Rightarrow \angle \mathrm{EMB}=\angle \mathrm{MND} \left(\mathrm{Alternate} \mathrm{Exterior} \mathrm{Angles} \mathrm{are} \mathrm{equal}\right) \\ \mathrm{So} \\ \Rightarrow \angle \mathrm{MND}+\angle \mathrm{BMN}=180° \\ \mathrm{Now} \mathrm{bisectors} \mathrm{that} \mathrm{is} \angle \mathrm{BMN}=2\angle \mathrm{PMN} \mathrm{and} \angle \mathrm{MND}=2\angle \mathrm{MNP} \\ \Rightarrow 2 \angle \mathrm{PMN}+2\angle \mathrm{MNP}=180° \\ \Rightarrow \angle \mathrm{PMN}+\angle \mathrm{MNP}=\frac{180°}{2}=90°.......1 \\ \mathrm{Now} \mathrm{in} \mathrm{triangle} \mathrm{MNP} \\ \Rightarrow \angle \mathrm{PMN}+\angle \mathrm{PNM}+\angle \mathrm{MPN}=180° \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{angles} \mathrm{in} \mathrm{triangle} \mathrm{is} 180° \\ \mathrm{From}1 \\ \Rightarrow 90°+\angle \mathrm{MPN}=180° \\ \Rightarrow \angle \mathrm{MPN}=90° \mathrm{Hence} \mathrm{proved} \end{gathered}$$

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