 Hello and welcome to the session. In this session we discussed the following question which says proof that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent. Before moving on to the solution, let's recall a fact which says that the perpendicular bisector of the chord of a circle always passes through. This is the key idea to be used for this question. First let's try to prove this key idea that we have just stated that the perpendicular bisector of the chord of a circle always passes through the center. Consider this chord AB of a circle with center O. Next we consider this DE to be the perpendicular bisector of the chord AB. Now we suppose that the perpendicular bisector of chord AB that is DE does not pass through the center of the circle that is O. So we have made this supposition. Now since DE is the perpendicular bisector of AB, so we have AD is equal to DB and also angle ADE is equal to 90 degrees. Now in the next step we join OD, so we have joined OD. Now we have also got that D is the midpoint of the chord AB and we know that the line joining the center of a circle to the midpoint of a chord is always perpendicular to the chord. This would mean that OD is perpendicular to the chord AB. Therefore we have angle ADO is equal to 90 degrees. Now angle ADE is equal to 90 degrees and angle ADO is equal to 90 degrees. So they both are equal. Angle ADE is equal to angle ADO. But this is a contradiction since angle ADO is a part of angle ADE. So this means our supposition that DE does not pass through O is wrong. Hence we get the perpendicular bisector DE of the chord AB passes through O that is the center of the circle. So we have proved the fact that the perpendicular bisector of the chord of a circle always passes through its center. Now let's see the solution of the given question. We take let ABCD be a cyclic quadrilateral and let O be the center of the circle which passes through the points ABC and D. So we get that AB, BC, CD and DA are the chords of the circle. Now from the result that we have stated that the perpendicular bisector of a chord of a circle always passes through its center. So we say that the perpendicular bisectors of the chords AB, BC, CD and DA pass through the point O which is the center of the circle. So this means that the perpendicular bisectors of the chords AB, BC, CD and DA of the circle are concurrent and these chords of the circle are the sides of the cyclic quadrilateral. Therefore the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent. So hence proved this completes the session. Hope you have understood the solution for this question.