**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* || *CD* and *AC* = 2 × *OD*.

**Solution:**

Given: *BC* is a diameter of a circle with centre *O *and *OD** *⊥* AB.*

To prove: *AC* parallel to *OD* and *AC = 2 × OD*

Construction: Join *AC*.

Proof:

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

Here, *OD *⊥* 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 ΔABC, 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., $O D \| A C$ and $O D=\frac{1}{2} A C$

∴ *AC = *2* × OD*

Hence, proved.

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