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  • AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters,

Q 7. AC and BD are chords of a circle that bisect each other. Prove that

(i) AC and BD are diameters

Answers (1)

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Given: AC and BD are chords of a circle which bisect each other.

To prove: AC and BD are diameters.

Construction: Join AB,BC,CD,DA. 

Proof :

In $\triangle \mathrm{ABD} \text { and } \triangle \mathrm{CDO}$,
$\mathrm{AO}=\mathrm{OC}$-----(Given)$
$\angle \mathrm{AOB}=\angle \mathrm{COD}$---- (Vertically opposite angles)
$\mathrm{BO}=\mathrm{DO}$----(Given)
So, $\triangle \mathrm{ABD} \cong \triangle \mathrm{CDO}$  (By SAS)
$\angle \mathrm{BAO}=\angle \mathrm{DCO} 
 ---(\mathrm{CPCT})$
$\angle \mathrm{BAO}$ and $\angle \mathrm{DCO}$ are alternate angles and are equal.
So, $A B \| D C$ ----------1
Also $A D \| B C$ ---------- 2

From 1 and 2,

$\angle A+\angle C=180^{\circ}$
$\angle \mathrm{A}=\angle \mathrm{C}$
From 3 and 4,
$\angle A+\angle A=180^{\circ}$
$\Rightarrow 2 \angle A=180^{\circ}$
$\Rightarrow \angle A=90^{\circ}$

BD is the diameter of the circle.

Similarly, AC is a diameter.

Posted by

seema garhwal

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