 Hello and how are you all today? My name is Priyanka. The question says if two equal chords of a circle intersect within the circle, proves that the segments of one chord are equal to the corresponding segments of the other chord. Now this is the figure that will help us in solving this question. We are given over here is that the chords are equal to each other. So AB is equal to CD and they are intersecting within the circle, so let us name this point as P. We need to prove that the segment of one chord that is AP and PB is equal to the corresponding segment of the other chord, so that will be PD is equal to PC. So we need to prove that AP is equal to PD that is the corresponding segments and PB is equal to PC or CP. So we need to construct draw OM perpendicular on AB, perpendicular on C, now let us start with our proof and also join OP, right? Now I know that perpendicular is the center to the chord by sixth chord, right? So that means that AM is equal to MB and similarly for the other chord also is equal to ND. So we are given that the chords are equal so that means their half will also be equal to each other, so therefore I can write that AM is equal to ND also MB is equal to CN, right? Since their halves are equal I can write that MB is equal to ND or CN anything and MB is also equal to one of the parts, one of the equal parts of the segment. So now I have in triangle OMP that is OMP, OMP I know that OM is equal to ON, the reason is that equal chords we are given that the chords are equal to each other AB is equal to CD. So equal chords are equidistant from the center, right? So the distance between their centers that is OM and ON are equal to each other. Further angle OMP is equal to angle OMP because they are both 90 degrees each and we know that the perpendicular from the center bisects the chord also and is perpendicular to the chord also and we have drawn it perpendicular itself and further OP is equal to OP that is the common part of the both triangles. So therefore triangle OMP is congruent to triangle OMP by RHS congruency criteria. Now I know that these two triangles are congruent to each other so their corresponding parts will also be equal to each other. So I can write that therefore MP is also equal to PN by CPCT. So this part is also equal to this part further proves that AM was equal to NT, it has been already proved above. So if I add M to both the sides, what I have over here is M plus MP is equal to ND plus MP also AM plus MP ND plus MP since it is equal to PN I can write PN in place of MP stating the reason because PM is equal to PN so now if you observe it AM plus MP is equal to what AP that implies AP is equal to ND plus will be equal to PD right so this proves our first part and further I know that AB is equal to CD so if I subtract P from both the sides I have AB minus AP is equal to CD minus AP now in place of AP what will I write exactly I will write that that implies AB minus this part will give us the rest of the portion that is PB equals to now in place of AP I can write down PD because AP is equal to PD proved above from this condition so that implies PB is equal to CD minus this portion will give us C right so these were the two statements that was required to be proved so we hence proved that the corresponding segments that is AP is equal to PD and PB is equal to PC that is the corresponding segments of the equal codes are equal to each other this completes my session hope you enjoyed it take care and bye for now