In the adjoining figure, OD is perpendicular to the chord AB of a circle whose centre is O. If BC is a diameter, then which of the following are true?

$A C = 2 O D$

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$\frac{A C}{D O} = \frac{B C}{B O}$

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$\frac{A C}{B O} = \frac{B C}{D O}$

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$O D ∥ A C$

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Solution

The correct options are
A
$A C = 2 O D$

B
$\frac{A C}{D O} = \frac{B C}{B O}$

D
$O D ∥ A C$

AB is the chord and OD $⊥$ AB $⟹$ BD = AD (Since, perpendicular from the centre to a chord bisects the chord)

In $△$ ABC, D is the mid-point of AB and O is the mid-point of BC.

$∴ O D ∥ A C$

Now, in $△$ ABC $O D ∥ A C$ .

Then by BPT, we must have

$\frac{A C}{D O} = \frac{B C}{B O}$ or $\frac{A C}{D O} = \frac{2 B O}{B O} = 2$

or $A C = 2 D O$

i.e., $A C = 2 O D$

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