Q. O IS THE CENTRE and AB is a chord. AC is the bisector of angle OAB .OAB = 56
a) prove that OC and AB are parallel
b) find angle ABC and OBE.

Dear Student,



$$\begin{gathered} Given : O is the centre and AB is a chord. AC is the bisector of \angle OAB which is 56°. \\ To Prove : OC\parallel AB \\ then \\ To Calculate : \angle ABC and \angle OBE. \\ Solution : Since AC is the angle bisector of \angle OAB=56° \\ \Rightarrow \angle OAC=\angle CAB=\frac{56}{2}=28° \\ Now, from arc \stackrel{⏜}{BC} \\ \Rightarrow \angle BOC=2\angle CAB (Angle subtended by an arc on centre is double the angle subtended on circumference) \\ \Rightarrow \angle BOC=2\times 28°=56° \\ Now, \\ OA=OB (Radius of the circle) \\ so, △OAB is isosceles \\ with OA=OB \\ so, \\ \angle OAB=\angle ABO=56° \\ Now, \sin ce, \angle ABO=\angle BOC=56° \\ \Rightarrow Alternate interior angles are equal \\ \Rightarrow \boxed{ OC\parallel AB } \\ Now, in △OBC \\ OB=OC\left(Radius\right) \\ \Rightarrow \angle OCB=\angle OBC \\ \Rightarrow \angle BOC+2\angle OBC=180° \\ \Rightarrow 56°+2\angle OBC=180° \\ \Rightarrow 2\angle OBC=180° -56° \\ \Rightarrow 2\angle OBC=124° \\ \Rightarrow \angle OBC=\frac{124°}{2} \\ \Rightarrow \angle OBC=62° \\ Now, \\ \angle OBA+\angle OBC+\angle CBE=180° (angles on a straight line) \\ \Rightarrow 56°+62°+\angle CBE=180° \\ \Rightarrow \angle CBE=180°-56°-62° \\ \Rightarrow \angle CBE=62° \\ Now, \\ Required angles \\ \angle ABC=\angle ABO+\angle OBC=56°+62°=118° \\ \angle OBE=\angle OBC+\angle CBE=62°+62°=124° \end{gathered}$$

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