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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