 In this video, we are going to learn about alternate segment theorem now the statement for alternate segment theorem is a little complicated So, let me just quickly give you an overview what we are going to do in the alternate segment theorem Basically, what we have is a chord and a tangent So whenever you see a triangle formed like this by taking a chord as a side of the triangle and the vertex of the Triangle is basically the point where the tangent that is the circle That's where you can assume that it could be the alternate segment theorem We will come to this statement of this theorem later What we really want to do here is to show that the angle BPN Like this is going to be equal to angle PAB Now I used to wonder often when in school. Why are we doing this? So all these proofs are like a chain So if you prove one thing you can prove another and from that another and then another and this is like a very Important result that we need to arrive at so this is just another proof that we need to study in order to reach Another interesting result in geometry. So let's not worry about it We want to prove that angle BPN is equal to angle PAB or you could also say we need to prove angle APM Like this to be equal to angle ABP Something like this now. We will only focus on one pair that we need to prove So let us first draw the center of this circle Let us assume that this is the center of the circle and I want to draw the radius So this is the radius and this is another radius now Let me just name this as oh, so I have the radii as OP and OB now Let us see if we can reduce something since OP is the radius. We can say that OP n is a right angle like this since OP is perpendicular to Mn because OP is the radius and Mn is a tangent and P is the point where radius and tangent meet and therefore We know that OP is perpendicular to Mn and therefore angle OP n 90 degrees This is a very important result for us right now. We will use it later. Also OP is congruent with OB OP is congruent with OB and therefore their lengths are equal. So OP is equal to OB The reason for this is that both are radii in triangle OPB since OP is equal to OB angle OPB is congruent with Angle OB P since the sides opposite to these angles So what I'm talking about this small angle here and small angle here OPB and OB P And the reason for this is that it's the property of an isosceles triangle Right, let me just write isosceles triangle property So this is the first point that I made is the second point that I made and this is third point that I'm talking about right now Now we already know that OP n is equal to 90 degrees to simplify Let me just give OPB and OB P some variable measure as X degrees So we'll use it later and we know another result since let me just put another point here so that I can Mention the name of the arc here. So let me just write D here. So arc PDB arc PDB Attends angle P O B at the center and Attends angle P AB at the circle We know the relationship between angle P O B and P AB because these are the angles Subtended at the center and at the circle by the same arc the relationship is as follows angle P O B is Twice as much as angle P AB is right now We can write relationship between the angles in the triangle P O B. So angle P O B Plus angle OP B plus angle OB P which are like X degrees each So let me just write X degrees plus X degrees to be equal to 180 degrees and therefore that means angle P O B is equal to 180 degrees minus 2x degrees we want the relationship between angle B P N and angle P AB We just saw that angle P O B is twice as angle P AB and therefore angle P AB is equal to half of angle P O B, which is 180 degree Minus 2x and that gives me a measure of angle P AB as 90 degrees minus X degrees now let us I want to highlight this and I will go back to the fact When we wrote angle O P N to be equal to 90 degrees I already know that angle O P N is equal to angle O P B plus angle B P N Angle O P B is equal to X degrees. So can I write the relationship as angle O P N? equal to angle O P B plus angle B P N now angle O P B is nothing but X degrees and so I can write X there and Angle O P N is in fact 90 degrees that we know. So I'll just write 90 degrees here And so I have the relationship which I can rearrange to write that angle B P N is equal to 90 degrees minus X degrees and I want to highlight this as well and So angle P AB is also 90 minus X and angle B P N is also 90 minus X and Because of these two results since the right-hand sides are equal I can say that angle B P N and angle P AB are equal So this was a purely mathematical proof now. Let us look at the statement of this theorem The statement says that measure of an angle formed by a tangent So this is the tangent to the circle and a chord through the point of contact So this is the tangent Mn and AP or BP could be considered as the chords through the point P And we are looking at the angle formed Between the chord and the tangent through the point of contact the point of contact is P here And so there are two angles possible that we could be talking about So one of the angles is BP N another is APM now this Angle is equal to the measure of an angle inscribed in the alternate arc angle B P N is formed in the arc BDP and so the alternate arc to this arc is arc P AB and the angle formed in it is angle P AB or B AP and We are talking about the angle in the alternate arc and the angle Formed by the chord and the tangent through the point of contact in the alternate arc. So therefore With this result we can write that angle BP N which we already shown is congruent with angle P AB and Also, I can write that Angle APM would be equal to the angle in the alternate arc Which is angle ABP and this is how we can conclude the result for Alternate segment theorem. What is a segment? So if I draw a circle like this and if I draw a chord one is a major segment and another is a minor segment Let me fill it up for you. So this blue is a major segment and pink is a minor segment And that's why then in the name of this theorem. There is the word segment So this is a major segment. Just remember major segment and this is a minor segment of a circle Don't confuse it with the normal line segment that you see and this is how we have proved the alternate segment theorem