 Hello and welcome to the session. In this session we discussed the following question which says in the given figure O is the center of a circle in which cods AB and CD intersect at P such that PO bisects angle BPD prove that AB is equal to CD. Before moving on to the solution let's recall one fact which says that the cods equidistant from the center of a circle are equal. This is the key idea that we use in this question. Now let's move on to the solution. This is the figure given to us in this we have that the cods AB and CD of the circle intersect at the point P then we are given this OP bisects angle BPD that is we have angle OPD is equal to angle OPD and we need to prove that AD is equal to CD. For this first we will draw OE perpendicular to the cod AB and OS perpendicular to the cod CD. So this OE is perpendicular to AB, OF is perpendicular to CD. So this means we now get angle OFP is equal to 90 degrees and angle OEP is equal to 90 degrees. We consider the triangles OFP and OEP in this angle OFP is equal to angle OEP equal to 90 degrees then the side OP is equal to OP that is the common side to both the triangles and then we were given that angle OPD is equal to angle OPB that is angle OPF is equal to angle OPE. So therefore we get triangle OFP is congruent to the triangle OEP this is by the AAS congruence rule. So this means that OF would be equal to OE that is the corresponding parts of congruent triangles are equal. So therefore we say that the cods AB and CD of the circle are equidistant from the center of the circle. Now in the key idea we stated the result which says that cods equidistant from the center of a circle are equal now that we have got cods AB and CD of a circle are equidistant from the center of the circle. So this means that both the cods AB and CD are equal and we were supposed to prove this so hence proved. So this completes the session hope you have understood the solution for this question.