$A, B, C$ are point in the circle with centre $O$ . If $O C A = x$ then find $∠ O A C$ . Prove that $∠ O C A$ + $∠ A B C = 90^{o}$ .

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Solution

$\begin{matrix} ∴ ∠ A O C = 2 ∠ A B C (b y p r o p e r i t i e s) a n d, ∠ O C A = ∠ O A C = x [O A = O C] N o w i n Δ A O C ∠ A O C + ∠ O A C + ∠ O C A = 180^{\circ} \Rightarrow 2 ∠ A B C + x + x = 180^{\circ} \Rightarrow 2 (∠ A B C + x) = 180^{\circ} \Rightarrow ∠ A B C + ∠ O A C = 90^{\circ} . p o v e d . \end{matrix}$

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