 Hello and welcome to the session. In this session, we will discuss the alternate segment property. First of all, let us see what are the segments of a circle. A chord divides the circle into two parts which are called segments of the circle. Consider this circle with center O and it has a chord AB. As you can see this chord AB divides the circle into two parts. One is the smaller part and one is the larger part and so these two parts are the segments of the circle. The smaller part is the minor segment of the circle and the bigger part or the larger part is the major segment of the circle. Consider this tangent PT to the circle as AB is the chord of contact. This is the major segment that is AXB is the major segment, AYB is the minor segment. You can see angle BAT segment AXB lie on the opposite sides of the chord of contact AB. Now if you consider the angle BAT then we say that AXB is the segment BAT. In the same way if you consider the angle BAP that is this angle then we say that AYB is the alternate segment for the angle BAP. The minor segment in this case which is AYB is the alternate segment for the angle BAP. Angle BAT AXB is the alternate segment and so this angle that is XB is the in the alternate segment. In the same way for angle BAP that is this angle AYB is the alternate segment and so this angle that is angle AYB is the alternate segment. Now we will discuss the theorem of the alternate segment or you can say the alternate segment property. According to which we have that is from the point of contact between the tangent effectively equal to the angles in the alternate segments. Considering this figure in which we have a circle with center O this P and Q is the tangent to the circle. From this point of contact of the tangent to the circle that is the point N we have drawn a chord MN. This chord divides the circle into two parts that is the segments of the circle. MSN is the major segment of the circle and MRN is the minor segment of the circle. Now if you consider the angle MNQ that is this angle the alternate segment for the angle MNQ is MSN. For the angle PMN that is this angle then for this angle the alternate segment is MRN. We have to prove that the angles between the tangent and the chord are respectively equal to the angles in the alternate segments. Where is Rossi? What all is given to us? We are given that PMQ tangent the circle has to prove the tangent and the chord. First we have angle NMQ is equal to the angle in the alternate segment. For the angle NMQ the alternate segment is and the segment MSN is this angle that is angle. So we have to prove that angle NMQ is equal to angle MSN. Then next we have to prove that this angle the angle in the alternate segment which is angle MRN. Structural. Structural. Now we will start with the proof. We will prove MQ is equal to angle at the tangent the radius perpendicular to each other. Therefore we can say that the radius OM is perpendicular to MQ. This OM is perpendicular to PQ so we can also say that OM is perpendicular to MQ. So this means that angle OMN that is this angle MQ that is this angle is equal to 90 degrees. Or we can say that angle TMN are equal so angle TNN plus angle NMQ is equal to 90 degrees. Let this be the result 1 ST is equal to 90 degrees. This is the angle in a semicircle and we know that angle in a semicircle is a right angle. Now as angle MST is equal to 90 degrees. So from this figure we can say that angle MSN angle MST is equal to 90 degrees. That is in place of angle MST we can write angle MSN plus angle MST. That is we add these two angles. Let this be the result. From the results 1 MN plus angle MNQ is equal to angle MSN plus angle MST. In the figure we find that angle ST that is equal to the angle that is this angle. So in the segment MSN we have the angles TMN and so they are equal. Now since these two angles are equal so they can cancel each other. Therefore we have angle MNQ is equal to angle MSN. This is what we were supposed to prove that we get that this angle between the tangent and the chord is equal to the angle in its alternate segment. So now let's prove the first part. Next we are supposed to prove is equal to the figure we find that MRNL quadrilateral is equal to 180 degrees. There are the opposite angles quadrilateral is equal to 180 degrees. Let this be result 3. Now again from the figure we have angle is equal to 180 degrees. Let this be result 4. The result that angle plus angle MSN is equal to angle plus angle PN. You already know that angle or you can say MNQ is equal to angle and we have proved part 1 that these two angles will cancel each other. We get angle MRN is equal to angle PNM is equal to angle. This is what we were supposed to prove. So as we have proved the second part also, proof of the alternate segment property. So this we can conclude which is on the circumference through that point is equal to the angle its segment tangent to the circle. We have that the straight line PNQ takes an MQ with the chord MN and this angle MMQ is equal to the angle in the alternate segment which is MSN. Hope you understood the alternate segment property.