A) a rectangle
B) a rhombus
C) a square
D) concyclic
Correct Answer: D
Solution :
OA \[\bot \] AP and OB \[\bot \] BP \[\angle \]OAP = \[90{}^\circ \] and \[\angle \]OBP = \[90{}^\circ \] \[\therefore \] \[\angle \]OAP + \[\angle \]OBP = \[90{}^\circ \] +\[90{}^\circ \]= \[180{}^\circ \] In quadrilateral OAPB, \[\angle \]OAP + \[\angle \]APB + \[\angle \]AOB+\[\angle \]OBP=\[360{}^\circ \] \[\Rightarrow \]\[\angle \]APB + \[\angle \]AOB = \[180{}^\circ \] \[\therefore \] The quadrilateral will be cyclic.You need to login to perform this action.
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