In the figure,OD is perpendicular to chord AB | vedclass

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(N/A) We have a circle with centre $O$. $BC$ is a diameter and $AB$ is a chord such that $OD \perp AB$. Join $AC$.The perpendicular from the centre of

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(N/A) We have a circle with centre $O$. $BC$ is a diameter and $AB$ is a chord such that $OD \perp AB$. Join $AC$.
The perpendicular from the centre of a circle to a chord bisects the chord. Therefore,$D$ is the midpoint of $AB$.
Since $O$ is the centre of the circle,$O$ is the midpoint of the diameter $BC$.
In $\triangle ABC$,$O$ is the midpoint of $BC$ and $D$ is the midpoint of $AB$. Therefore,$OD$ is the line segment joining the midpoints of two sides of $\triangle ABC$.
By the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of it.
Therefore,$OD \parallel CA$ and $OD = \frac{1}{2} CA$.
Hence,$CA = 2OD$.

In the figure,$OD$ is perpendicular to chord $AB$ of a circle whose centre is $O.$ If $BC$ is a diameter,prove that $CA = 2OD$.

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