In the given Figure, circle $C (O, r)$ and $C (O^{'}, r / 2)$ touch internally at a point $A$ and $A B$ is a chord of the circle $C (O, r)$ intersecting $C (O^{'}, r / 2)$ at $C$ . Prove that $A C = C B$ .

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Solution

Join $O A, O C$ and $O B$ .

Clearly, $∠ O C A$ is the angle in a semi-circle.

$∴ ∠ O C A = 90^{o}$

In right triangles $O C A$ and $O C B$ , we have

$O A = O B = r$

$∠ O C A = ∠ O C B = 90^{o}$ and, $O C = O C$

So, by $R H S -$ criterion of congruence, we get

$△ O C A ≅ △ O C B$

$∴ A C = C B$

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In the given Figure, circle C(O,r) and C(O′,r/2) touch internally at a point A and AB is a chord of the circle C(O,r) intersecting C(O′,r/2) at C. Prove that AC=CB.

Join OA,OC and OB.Clearly, ∠OCA is the angle in a semi-circle.∴ ∠OCA=90oIn right triangles OCA and OCB, we haveOA=OB=r∠OCA=∠OCB=90o and, OC=OCSo, by RHS−criterion of congruence, we get△OCA≅△OCB∴AC=CB

In the given Figure, circle $C (O, r)$ and $C (O^{'}, r / 2)$ touch internally at a point $A$ and $A B$ is a chord of the circle $C (O, r)$ intersecting $C (O^{'}, r / 2)$ at $C$ . Prove that $A C = C B$ .