 Hello and welcome to the session. In this session, we discussed the following question which says proves that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. The second part of the question says in triangle PQR, the bisector of angle Q needs PR at S, a line Tt dash parallel to PR needs PQ, QS and QR at T, O and T dash respectively. Show that TO multiplied by QT dash is equal to T dash O multiplied by QT. Before we move on to the solution, let's recall one result which says that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. This is the key idea that we use for this question. Now let's see the solution. First let's try to prove the first part of the question in which we have to prove that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. For this, let's consider this triangle PQR. So we are given a triangle PQR in which this Pt is the bisector of angle P. That is we have angle 1 is equal to angle 2. We need to prove that the internal bisector Pt of the angle P of triangle PQR divides the opposite side QR internally in the ratio of the sides containing the angle P. That is we need to prove that QT upon TR is equal to PQ upon PR. For this we need to do some construction. So we draw RT dash parallel to TP meeting QP produced at T dash. So as you can see we have drawn RT dash parallel to TP. Now since we have RT dash is parallel to TP. This means angle 2 is equal to angle 3 since they are the alternate interior angles and also we have angle 1 is equal to angle 4 as they are the corresponding angles. But we also have angle 1 is equal to angle 2 since Pt is the bisector of angle P. So therefore we get angle 3 is equal to angle 4. Now in triangle PR T dash angle 3 is equal to angle 4. So this means PR is equal to Pt dash since we know that in a triangle sides opposite equal angles are equal. Now in triangle QT dash R we have Pt is parallel to T dash R and we have a result which says that if a line is drawn parallel to one side of a triangle to intersect the other two sides and distinct points then the other two sides are divided in the same ratio. So in triangle QT dash R Pt is parallel to T dash R. So therefore we have QT upon TR is equal to QP upon Pt dash. Now since PR is equal to Pt dash therefore we get QT upon TR is equal to QP or you can say PQ upon PR. So we were supposed to prove this and so hence proved. See the second part of the question in which we have that in triangle PQR the bisector of angle Q meets PR at S. So in this figure we have a triangle PQR the bisector of angle Q meets PR at the point S and a line T T dash is parallel to PR and it meets PQ QS and QR at T O and T dash respectively. We need to show that TO multiplied by QT dash is equal to T dash O multiplied by QT. So in this part of the question we are given a triangle PQR and we are given that QS is the bisector of angle Q. We also have that T T dash is parallel to PR. We need to show TO multiplied by QT dash is equal to T dash O multiplied by QT. Now as you know that in triangle PQR QS is the internal bisector of angle Q. So you can say that QO is the internal bisector of angle Q of triangle PQT dash. Now using this above result that we have just proved it says that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. We can say that as QO is the internal bisector of angle Q for triangle TQT dash TO upon T dash O is equal to QT upon QT dash. So now cross multiplying we get TO multiplied by QT dash is equal to QT multiplied by T dash O. And we were supposed to prove this only that is TO multiplied by QT dash is equal to T dash O multiplied by QT. So hence proved this completes our session. Hope you have understood the solution of this question.