 Hello and welcome to another session on jumps of geometry Now in this theorem, it's given that the external bisectors of any two angles of a triangle Concurrent with the internal bisector of the third angle so Points to be noted here First of all, they're saying external bisectors of any two angles of a triangle Let's try and understand what's external bisector is first of all. So let's say this is a triangle and If I produce one of the sides So a b c is a triangle and let's say I produce this bc side to D then if I draw this line which is Let's say cm Now if these two angles are equal, then we say that cm is the external angle bisector of ad now We can have, you know Multiple external angles at a point actually. So if I produce let's say a c Like that. So this could also be the external angle, isn't it? So let's say this is theta This is theta. So hence multiple external angles are possible For a triangle, okay now so once we have understood this what is an external back bisector Now we need to understand what x circle is. So let's understand what an x circle is So many of you would have come across the concept of in circle But let's understand what x circle is. So first of all, we have to have our triangle. So let me draw a triangle So this is a triangle and now I let's say this triangle is a bc and I extend Right bc and extend Let's say b a like that. Now guys The circle which is tangent to all the three sides which three sides B a extended, let's say this is d bc extended Let's say e and the side ac. So it would appear to be something like that Okay, so this circle which will be formed which is Tangent to all the three sides Yeah, so one ac and the other two b a extended and bc extended will be called as x circle and the center of this circle, let's say the center is over here wherein, let's say if you drop a perpendicular on a b like that and Like that on the three sides. Let's say like that. So these you know the center this o is the x center What is it x? Center, okay, and the circle this one here will be called x circle So you can imagine if this is one x circle, there could be you know couple of more in fact two more So if I draw this draw that and here is another circle like that So this one will be another one and let's say if you produce this and produce that So there will be another one like that Okay, so there could be three x circles and x centers possible and the one which is let's say inscribed within the Triangle will be called be in circle and the center of this will be called in Center so I hope you now understood the concept of x circle and in circle, right? Now let us explore the given theorem. It says the external bisectors of any two angles of a triangle, okay? So in the given picture if you see closely a bc is the triangle So let me highlight the triangle a bc is the given triangle, okay? and You can see I a or z a z b if you can notice this z a Z b Right, so basically a b is being produced and x c and x b So bc is being produced in both sides and Y c and y a so a b ac is produced on both sides. So if you see You have external angles z a So this z a bc this one is an external angle, isn't it this angle is an external angle and it's bisector I see I be so this line here is the bisector Okay off this external angle similarly I a See this one is an external bisector of Angle b c y a angle b c y a So you can take your time and see b c y a so these are the external bisectors and The question our theorem says the external bisectors of any two angles of a triangle are concurrent With the internal bisector of the third angle. So these are the two external bisectors. I a b and I a c are the two external bisectors and we have to prove that a I a a I a is the internal bisector of angle B a c a I a is bisector of Angle B a C Right, this is what the theorem is suggesting similarly if you take these two external bisectors then B Angle the bicep internal bisector of B that is this line B. I B Also pass through I B or you have to prove that if you join B. I B B and I B then it is the Internal bisector or bisector of internal angle B and likewise the other side this one So these are the two external bisectors and Right, so I hope you understood the theorem. Okay, so now let's look at the proof so I a B I a B is the External Or sorry not external is the bisector. I a B is the bisector of Angle I a B C Sorry not I a Z a Z a Z a B C Z a B C Okay, so I a B is the bisector of angle Z a B C and I A C I a C is the bisector of Angle Z sorry why a Why a C B Correct. Look closely. You will get it now. That means I a is equidistant equidistant from From which two sides B Z a and B C Why why is that because we know that if Because angle bisector is nothing but the locus of Points which are equidistant from the two sides and we have seen this in the previous session So hence if this is an angle and if you trace the trace the locus of all the points which are equidistant from both the Sides then you will get the angle bisector. So this will be angle bisector. This is what we have learned, isn't it? So I a is equidistant from B Z a and B C similarly I a is equidistant equidistant from BC and and Why a C because if you see this point I a is lying on the angle bisector this angle bisector then I a must be equidistant from both from both DC as well as Why a C is it now from these two we can say I a I a is equi equidistant from Which two sides B Z a as well as why a C? correct I a is equidistant from B Z a and why a C or I a is equidistant equidistant from I can say a Z a a a Z a and a Why a isn't it because if you see B Z a is nothing but extended a Z a or Extended a B is B Z a and extended a C is a why a okay now from this We can say What what can we say about it so hence? So I a is equidistant from B Z a And why a C or I a is equidistant from a Z a and A why a now what is this if you see clearly that means I a must I a must lie on the bisector of angle Z a a Why a isn't it because if a point is equidistant again the same theorem if a point is equidistant from two intersecting lines Right two intersecting lines. They are equal as this is equal to this then this point will lie on the angle bisector why? because if you take these two triangle, let's say P Q R and S in these two triangles these two if you see these two triangles what happens? So this side or let's say this side is what this side This is 90 degrees and this side is common. That means these two triangles are congruent and hence This angles would be equal. Okay, so hence we prove that I a must lie I a must lie on the bisector of angle Z a A why a and what is it? It is nothing but I a lies on bisector I a lies on bisector of angle B a C hence Proved right that means all the three bisectors two external bisectors External angle bisector and one internal third angle internal bisector all are Concurrent so this is a very interesting theorem