 Hello and welcome to the session. My name is Mansi and I am going to help you with the following question. The question says A, B, C, D is a parallelogram. The circle through A, B and C intersect C, D produced if necessary at E. Prove that A equal to E, D. We have to prove that this side is equal to this side. So let us start with the solution. Now in order to prove that A, E is equal to A, D, it is sufficient to prove that angle A, E, D is equal to angle A, D, E because if we prove that these two angles are equal then sides opposite to them will be also equal. Since A, B, C, E is a cyclic quadrilateral therefore we can say that angle A, E, D plus angle A, B, C is equal to 180 degree. We call this one, this happens because we know that in a cyclic quadrilateral the sum of opposite angles is equal to 180 degree. Now C, D, E is a straight line. So now E is a straight line and because this is a straight line so this implies that angle A, D, E plus angle A, D, C will be equal to 180 degree because we see that since this is a straight line so sum of linear angles will be equal to 180 degree we call this two. From one and two we have, one and two we have angle A, E, A, B, C is equal to angle A, D plus angle A, B, C because both of them they are equal to 180 degree. This implies that here angle A, B, C gets cancelled with angle A, B, C and we have angle A, E, D is equal to angle therefore angle A, E, D we have is equal to angle A, D, E and since these two angles are equal so sides opposite to them will be also equal and this implies that A, D is equal to A, E. Now this was what we were supposed to prove in this question. So I hope that you understood the question and enjoyed the session. Have a good day.