Answer :

The figure is given below. O1 and O2 are the centres of the big and the small circles respectively.


Now, AB is the diameter of big circle.


AO1 is the diameter of the small circle.



Construction:


Let us join O1C and DB. The constructed image:



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


Hence, ACO1 = 90° And, ADB = 90°


Now, in ΔACO1 and ΔADB,


A is common to both the triangles.


AO �1 = BO1 (the radii of the big circle)


ACO1 = ADB = 90°


Hence,


in ΔACO1 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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