Answer :


Assertion (A):


Construction: Draw a Δ ABC in which AB = AC, Let O be the midpoint of AB and with O as centre and OA as radius draw a circle, meeting BC at D


Now, In Δ ABD


ADB = 90° (angle in semicircle)


Also, ADB + ADC = 180°


90° + ADC = 180°


ADC = 180° – 90°


ADC = 90°


Consider Δ ADB and Δ ADC


Here,


AB = AC (given)


AD = AD (common)


ADB = ADC ( 90° )


By SAS congruency, Δ ADB Δ ADC


So, BD = DC(C.P.C.T)


Thus, the given circle bisects the base. So, Assertion (A) is true


Reason (R) :


Let BAC be an angle in a semicircle with centre O and diameter BOC


Now, the angle subtended by arc BOC at the centre is BOC = 2× 90°


BOC = 2× BAC = 2× 90°


So, BAC = 90° (right angle)


So, reason (R) is true


Clearly, reason (R) gives assertion (A)


Hence, correct choice is A

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