Answer :

The figure is given below. O_{1} and O_{2} are the centres of the big and the small circles respectively.

Now, AB is the diameter of big circle.

AO_{1} is the diameter of the small circle.

__Construction:__

Let us join O_{1}C and DB. The constructed image:

Now, since the diameter subtends a right angle in the circle at any point.

Hence, ∠ACO_{1} = 90° And, ∠ADB = 90°

Now, in ΔACO_{1} and ΔADB,

∠A is common to both the triangles.

AO �_{1} = BO_{1} (the radii of the big circle)

∠ACO_{1} = ∠ADB = 90°

Hence,

in ΔACO_{1}∼ ΔADB,

⇒

Since, the ratio is always equal, we can say that any chord of the larger circle through the point where the circles meet is bisected by the small circle.

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