In the given figure,if $\angle ABC = 45^{\circ}$,prove that $OA \perp OC$.
(N/A) We know that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle AOC = 2 \times \angle ABC$.
Given that $\angle ABC = 45^{\circ}$.
Substituting the value,we get $\angle AOC = 2 \times 45^{\circ} = 90^{\circ}$.
Since $\angle AOC = 90^{\circ}$,the lines $OA$ and $OC$ are perpendicular to each other.
Hence,$OA \perp OC$ is proved.
Therefore,$\angle AOC = 2 \times \angle ABC$.
Given that $\angle ABC = 45^{\circ}$.
Substituting the value,we get $\angle AOC = 2 \times 45^{\circ} = 90^{\circ}$.
Since $\angle AOC = 90^{\circ}$,the lines $OA$ and $OC$ are perpendicular to each other.
Hence,$OA \perp OC$ is proved.
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