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  • Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.         <div

Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.

 

 

 

 
 
 
 
 

Answers (1)

Given : BC is chord of circle. AB and AC are tangents. M is point of interaction of BC and OA.

To prove : \angle ABM = \angle ACM

As we know : OC=BR=r\; (\text{radius of circle})

In  \bigtriangleup OBC,

\angle OBM =\angle OCM \; \; \; (\because OB=OC=r)

We also know that\angle ABO =\angle ACO = 90^{\circ} \; \; \; (\text{angle between radius and tangent }= 90^{\circ})

Now 

\angle ABO =\angle ABM + \angle OBM\; \; \; \; -(i)

\angle ACO =\angle ACM + \angle OCM\; \; \; \; -(ii)

Substract eq (ii) from (i) :

\Rightarrow \angle ABO-\angle ACO =\angle ABM - \angle ACM +\angle OBM - \angle OCM

\Rightarrow 0 =\angle ABM - \angle ACM +0 \; \; \; \; \left ( \because \angle ABO = \angle ACO \; \; and \; \; \angle OBM = \angle OCM \right )

\Rightarrow \angle ABM = \angle ACM

Hence proved.

 

Posted by

Ravindra Pindel

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