Question

Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.

Answer 1

To prove : A diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle i.e , PQ bisects ∠AOB

 

Solution :  

Let PQ is a diameter of circle which bisects chord AB to C.

Now,

In ∆BOC and ∆AOC,  

OA = OB  

[Radius]

OC = OC  

[Common]

AC = BC  

[Given]

Then, by SSS congruence criterion , we obtain ΔAOC ≅ ΔBOC

So, ∠AOC = ∠BOC

[By CPCT]

∴ PQ bisects ∠AOB.  

Hence proved that PQ bisects ∠AOB.  

HOPE THIS ANSWER WILL HELP YOU…..

 

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Answer 2

Answer:

Step-by-step explanation:

In ∆BOC and ∆AOC,  

OA = OB  

[Radius]

OC = OC  

[Common]

AC = BC  

[Given]

Then, by SSS congruence criterion , we obtain ΔAOC ≅ ΔBOC

So, ∠AOC = ∠BOC

[By CPCT]

∴ PQ bisects ∠AOB.