Class 9

Math

All topics

Circles

In the adjoining figure $OD$ is perpendicular to the chord $AB$ of a circle with centre $O$. If $BC$ is a diameter, show that $AC∥DO$ and $AC=2×OD$

Perpendicular from the centre of a circle to a chord bisects the chord

We know that $OB⊥AB$

From the figure we know that $D$ is the midpoint of $AB$

We get

$AD=BD$

We also know that $O$ is the midpoint of $BC$

We get

$OC=OB$

Consider $△ABC$

Using the midpoint theorem

We get $OB∥AC$ and

$OD=21 ×AC$

By cross multiplication

$AC=2×OD$

Therefore, it is proved that $AC∥DO$ and $AC=2×OD$