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  • In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Q : 10     In any triangle ABC, if the angle bisector of  \small \angle A and perpendicular bisector of BC
               intersect, prove that they intersect on the circumcircle of the triangle ABC.

Answers (1)

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Given :In any triangle ABC, if the angle bisector of  \small \angle A and perpendicular bisector of BC  intersect.

To prove : D lies on perpendicular bisector BC.

Construction: Join BD and DC.

Proof :

           

Let \angleABD = \angle1 ,  \angleADC = \angle2 , \angleDCB =\angle3 , \angleCBD = \angle4

\angle1 and \angle3 lies in same segment.So,

        \angle1 = \angle3    ..........................1(angles in same segment)

similarly, \angle2 = \angle4  ......................2

also,       \angle1=\angle2 ..............3(given)

From 1,2,3 , we get

     \angle3 = \angle4

Hence,  BD = DC (angles opposite to equal sides are equal )

All points lying on perpendicular bisector BC will be equidistant from B and C.

Thus, point D also lies on perpendicular bisector BC.

 

 

 

 

 

        

 

    

 

 

 

Posted by

seema garhwal

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