 Hello and welcome to the session. In this session, we discuss the following question that says P is the midpoint of site AB of the parallelogram ABCD. Aligned through B, parallel to PD, needs BC at Q and AD produced at R. Proof that AR is equal to 2 times BC and BR is equal to 2 times BQ. Before moving on to the solution, let's state the converse of the midpoint theorem. According to this, we have that a line through the midpoint of a site of a triangle parallel to another site by 60 third site. This is the key idea for this question. Now we move on to the solution. Consider this figure. In this, we are given that ABCD is a parallelogram. It's given that P is the midpoint of site AB. That is we have AP is equal to PB. A line is drawn through the point B parallel to PD and it needs DC in Q and AD produced at R. That is we have BR is parallel to PD or we can also say BQ is parallel to PD. We need to prove AR is equal to 2 times BC and BR is equal to 2 times BQ. First we shall prove AR is equal to 2 times BC. Now we consider triangle ARB in this. P is the midpoint of the site AB. Now PD is parallel to BR. It's already given to us. So a line through the midpoint of one site of the triangle ARB is drawn parallel to the third site. So by converse of the midpoint theorem we have this line from P. That is the line PD would by 60 third site AR of the triangle ARB. That is we now say AD is equal to BR by the converse of midpoint theorem. Now since we have ABCD is a parallelogram. Therefore its opposite sites would be equal and hence we have AD is equal to BC. Now from the figure you can see that AR is equal to AD plus DR and we know that AD is equal to DR. So AR would be equal to 2 times AD and since AD is equal to BC. Therefore we get AR is equal to 2 times BC. So we have proved this AR is equal to 2 times BC. Next we need to prove that BR is equal to 2 times BQ. For this also we consider the triangle ARB. Now in the first part we had shown that AD is equal to DR. That is we have D is the midpoint of the site AR of triangle ARB. And since we have that ABCD is a parallelogram. Therefore CD is parallel to AB or we can say that DQ is parallel to AB. Now D is the midpoint of site AR of triangle ARB and DQ is parallel to AB. So by converse of midpoint theorem we have Q is the midpoint of the site RB. That is QR is equal to QB. Now from the figure we have BR is equal to BQ plus QR. Now BQ or say QB is equal to QR. Therefore we get BR is equal to 2 times BQ. And so we have proved this part also. So hence proved. This completes the session. Hope you have understood the solution for this question.