Question

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 $AC=2\times OD$.

Answer 1

Given: $BC$ is a diameter of a circle with centre $O$ and$OD\perp AB$.
To prove: $AC$ parallel to $OD$ and $AC=2\times OD$
Construction: Join $AC$ .

Proof:
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
Here, $OD\perp AB$
$D$ is the mid point of $AB$.
i.e., $AD=BD$
Also, $O$ is the mid point of $BC$
.i.e., $OC=OB$
Now, in $\Delta ABC$, we have:
$D$ is the mid point of $AB$ and $O$ is the mid point of $BC$.
According to the mid point theorem, the line segment joining the mid points of any two sides of a triangle is parallel to the third side and equal to half of it.
i.e., $OD\parallel AC$ and $OD=\frac{1}{2}AC$
$\Rightarrow AC=2\times OD$
Hence, proved.