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Standard X

Mathematics

Theorem of Equal Chords Subtending Equal Angles at the Center

In fig., circ...

Question

In fig., circles C(O,r) and $C (O^{'}, \frac{r}{2})$ touch internally at a point A and AB is a chord of the circle C (O,r) intersecting $C (O^{'}, \frac{r}{2})$ at C. Prove that AC = CB. [3 MARKS]

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Solution

Concept: 1 Mark
Application: 2 Marks

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

OA = OB = r

and OC = OC

Join OA, OC and OB .

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

So, by RHS criterion of congruence, we get

$Δ O C A$ is similar to $Δ O C B$

$\Rightarrow A C = C B$

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In fig., circles C(O,r) and C(O′,r2) touch internally at a point A and AB is a chord of the circle C (O,r) intersecting C(O′,r2) at C. Prove that AC = CB. [3 MARKS]

Concept: 1 Mark Application: 2 Marks In right triangles ΔOCA and ΔOCB, we have OA = OB = r and OC = OC Join OA, OC and OB . Clearly, ∠OCA is the angle in a semi circle . So, by RHS criterion of congruence, we get ΔOCA is similar to ΔOCB ⇒AC=CB

In fig., circles C(O,r) and $C (O^{'}, \frac{r}{2})$ touch internally at a point A and AB is a chord of the circle C (O,r) intersecting $C (O^{'}, \frac{r}{2})$ at C. Prove that AC = CB. [3 MARKS]