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In the adjoining figure OD is perpendicular to the chord AB of

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In the adjoining figure $O D$ is perpendicular to the chord $A B$ of a circle with centre $O$ . If $BC$ is a diameter, show that $A C ∥ D O$ and $A C = 2 \times O D$

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In the adjoining figure

O D

is perpendicular to the chord

A B

of a circle with centre

O

. If

BC

is a diameter, show that

A C ∥ D O

and

A C = 2 \times O D

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Text solutionVerified

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

We know that

OB ⊥ A B

From the figure we know that

D

is the midpoint of

A B

We get

A D = B D

We also know that

O

is the midpoint of

BC

We get

OC = OB

Consider

△ A BC

Using the midpoint theorem

We get

OB ∥ A C

and

O D = \frac{1}{2} \times A C

By cross multiplication

A C = 2 \times O D

Therefore, it is proved that

A C ∥ D O

and

A C = 2 \times O D

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Question Text	In the adjoining figure $O D$ is perpendicular to the chord $A B$ of a circle with centre $O$ . If $BC$ is a diameter, show that $A C ∥ D O$ and $A C = 2 \times O D$
Answer Type	Text solution:1
Upvotes	150

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

Text solutionVerified

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In the adjoining figure $O D$ is perpendicular to the chord $A B$ of a circle with centre $O$ . If $BC$ is a diameter, show that $A C ∥ D O$ and $A C = 2 \times O D$