Given: Two circles with centres O and O′, radii 2r and r respectively, touching internally at A. Chord AB of the bigger circle meets the smaller circle at C.
To prove: C bisects AB, i.e., AC = CB.
Proof:
Since the two circles touch internally at A, the point of tangency is A, and O, O′, A are collinear (the line joining the centres passes through the point of contact).
Join O′C.
In the smaller circle, O′A = O′C = r (radii of smaller circle).
Now, OA = 2r (radius of bigger circle), and OO′ = 2r − r = r.
Since AB is a chord of the bigger circle, join OC and drop a perpendicular — or use the following:
O′C = O′A = r, so triangle O′AC is isosceles. But more directly:
O′C ⊥ AB — because AB is a chord of the bigger circle, and O′ lies such that O′C is the radius of the smaller circle to point C, where the chord meets the smaller circle.
In the bigger circle, the perpendicular from the centre O to the chord AB bisects AB. Since O, O′, A are collinear and OO′ ∥ is along OA, we show O′C ⊥ AB:
The chord AB of the larger circle, when it meets the smaller circle at C, makes O′C a radius of the smaller circle. Since O′A = O′C = r, triangle O′AC is isosceles.
The line OA (i.e., line OO′A) is perpendicular to the chord AB at its midpoint — but the perpendicular from O′ to chord AB of the smaller circle (chord AC) bisects AC only if O′C ⊥ AB.
Correct approach:
O′C ⊥ AB: Since AB is a chord of the larger circle and O′C is the radius of the smaller circle at C, by Theorem 10.1 (tangent ⊥ radius), the smaller circle is internally tangent — instead, note:
The perpendicular from O′ on chord AB of the smaller circle (chord AC, since A and C lie on smaller circle) bisects AC. But we need C to bisect AB.
Key step: In the larger circle, OB is a radius (OB = 2r). Join O′C; O′C = r. Since OO′ = r and O′C = r, and OC = 2r (if C lies on the larger circle — but C lies on the chord, not necessarily on larger circle).
Draw O′M ⊥ AB. In the smaller circle, A and C lie on it, so O′M bisects AC → AM = MC.
In the larger circle, the perpendicular from O on AB: since OO′ is along OA, and O′M ⊥ AB, then OM ⊥ AB too (same perpendicular line), so OM bisects AB → AM = MB.
From AM = MC and AM = MB, we get MC = MB, i.e., C bisects AB. $\blacksquare$
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