Please slove question 18

​Q18. Two circles of equal radii cut each other at P and Q, so that the centre of one lies on the other. A straight line through P cuts the circle again at A and B. Prove that $∆$QAB is equilateral triangle.

Dear Student,

Please find below the solution to the asked query:

We form our diagram from given information , As :

Here ' A '  and ' B '  are centers of given circles

And

AQ =  BQ  =  AB = Radius of given circles (  As given both circles have same radius )

And

In $∆$ QAB , we know :  AQ =  BQ  =  AB , So we can say that

$∆$ QAB is an equilateral triangle .                                                         ( Hence proved )

Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards