In a circle with centre O, $O D ⊥$ chord AB. If BC is the diameter, then

A C = B C

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O D = B D

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A C = 2 O D

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none of these

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Solution

The correct option is C $A C = 2 O D$
$∠ O D B = ˚ 90 (O D ⊥ A B)$
$∠ C A B = ˚ 90$ ( Angle in a semi circle is 90 degree)
$I n △ B O D a n d △ B C A w e h a v e$
$∠ O D B = ∠ C A B (˚ 90)$
$∠ O B D = ∠ C B A$ (COMMON)
$∴ △ B O D \sim △ B C A (A A s i m i l a r i t y)$
$\frac{O B}{B C} = \frac{O D}{C A} = \frac{B D}{A B}$ ( Corrosponding sides are proportional)
$\frac{C A}{O D} = \frac{B C}{O B}$
$\frac{C A}{O D} = \frac{2 O B}{O B}$
$\frac{A C}{O D} = 2$
$A C = 2 O D$

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