Question

Can you Solve the marked question?

Answer 1

Step-by-step explanation:

In parallelogram PQRS:

$\angle SPQ + \angle RQP = 180\degree \\ \therefore \: \frac{1}{2} \angle SPQ + \frac{1}{2}\angle RQP =\frac{1}{2} \times 180\degree \\ \therefore \: \frac{1}{2} \angle SPQ + \frac{1}{2}\angle RQP = 90\degree.....(1) \\ \because \\ In \: parallelogram \: PQRS, \: \\ PO \: and \: RO \: are \: the \: angle \: \\ bisectors \: of \: \angle \: P \: and \: \angle \: Q \: \\ respectively. \\\therefore \angle OPQ= \frac{1}{2} \angle SPQ..... (2)\\</p><p>and\: \angle OQP= \frac{1}{2} \angle RQP......(3)\\From\: equations\: (1), (2) \:& \: (3)\\</p><p>\fbox {\angle OPQ+ \angle OQP= 90\degree} ....(4)\\</p><p>In\: \triangle\: OPQ\\</p><p>\fbox{\angle OPQ+ \angle OQP} + \angle POQ= 180\degree\\</p><p>90\degree + \angle POQ= 180\degree\\</p><p>\implies \angle POQ= 180\degree-90\degree\\</p><p>\implies \angle POQ= 90\degree$

Thus proved.