 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 the circle and X or Y is the diameter. If XZ is any other chord of the circle, show that XY is greater than XZ. Before we move on to the solution, we should know that sum of any two sides of a triangle is greater than the third side. This is the key idea to be used in this question. Now we move on to the solution. This is the figure given to us in which we are given that X or Y is the diameter of the circle, XZ is any chord of the circle and we need to prove that XY is greater than XZ. First we consider triangle XOZ. In this, we will consider the sum of the two sides of this triangle that is the sum of the sides OX and OZ and this would be obviously greater than the third side that is XZ. Since we know that sum of any two sides of a triangle is greater than the third side. Now this would further give us OX plus OY is greater than XZ. Since we have OX is equal to OY is equal to OZ since they are the radii of the circle. So instead of OZ, we have written OY. Now from the figure you can see that OX plus OY is equal to XY so we have XY is greater than XZ. So we have proved that XY is greater than XZ hence proved. So this completes the session. Hope you have understood the solution for this question.