 session on triangles and we are going to take up another validation for one more theorem now what is this theorem so there are two parallel lines AB and CD okay and FG or small h happens to be a transversal and you can see that this HF is the bisector of angle CFG FI is the bisector of DFG HG is the bisector of AGF and GI is the bisector of FGB all these bisectors are meeting one pair of them are meeting at H and the other one other ones are meeting at I now the theorem says that whenever there are two parallel lines and there's a transversal and you join the you know or enclose a quadrilateral with the you know the bisectors of the four interior angles then the quadrilateral so enclosed will always be a rectangle and I have shown that as 90 degrees you can see all the four angles are 90 degrees this actually is true and before proving this particular theorem we will as we have been doing we will be proving or validating it right you know changing the configuration of a parallel line and see that every time it is actually so so let's try and see if I move the point A and along with the direction of the parallel lines are going to move so this is the new direction new orientation so again in this case the quadrilateral enclose is a rectangle now let me see if I get any such case where it is not a rectangle so by all means you can see whether I change the position of C or D or any configuration I am trying to change all these points but every time you are getting a rectangle every time you are getting a rectangle right every time you are getting a rectangle all this is rectangle right now you might say how do you know that these are angle these angles are same so for one case I've just arbitrarily stopped it somewhere and now we'll see that yes indeed all these angles also are same so let me measure these angles so I'm measuring angle C F H and you can see that is forty nine point zero six and H F G forty nine point zero six both angles are same that means both the angles are same that means H F is indeed a bisector of C F G is it let me now show you some other angles so let's say now I'm trying to show you G F I this is forty point nine four and I F D this is also forty point nine four so that means these two are also equal angles which two angles I'm talking about the angle G F D is divided by IF into two equal parts both are forty point nine four so let me turn them off as well and show you the other ones so let's say I G and B so B G I forty nine point zero six and I G and F that is also forty nine point zero six so this is very much valid guys isn't it and if I now show you the last pair it will be same again so F G H forty one nine four and H G and A forty point nine four both are same angle so that means all these are perpendicular or sorry not perpendicular but angle bisector and I can change the orientation here as well and you can see all the angles still are perpendicular not perpendicular sorry angle are bisectors so these H G and F I and I G and F H all are bisectors of respective angles so in every case it is a rectangle that's what we are going to prove in the subsequent part of the session so we saw the validation of the theorem in the previous part of the video now we are going to take up the generalized proof so we are now going to prove that if parallel lines are intersected by a transversal then the bisectors of the two pairs of interior angles enclose or rectangle now what is given so given is AV parallel to CD okay and G F is a transversal right this is given what is to prove we have to prove that E F no G H F I or F I F I G H is a rectangle so what are the properties of rectangle we need to know so that we can prove it so clearly each of the angle must be 90 degree either that or you prove that if GI sorry not if F I G H is a parallelogram and then prove any one of the angle is 90 degree then also the rectangle is going to be sorry the quadrilateral is going to be a rectangle so what did I say I will write down either you prove all angles are right angles okay all angles are right angles then it is a rectangle or else parallelogram with one right angle with one right angle this also is good enough to prove that the given quadrilateral is a rectangle okay so in my opinion I will be going for the second case because it's easy to prove parallel in this case now let's prove this to you so clearly let me you know put some names on the angle so that becomes easier to identify and lesser effort in writing so let's say this is one this is two this is three this is four and this is five this is six this is seven and this is eight let's say these are the angle configuration meanwhile we have to also say that H F H G F I and IG are angle by sectors angle by sectors and which angle it's quite evident from the figure okay so we have to prove now that F I G H is a rectangle let's do that since AB is parallel to CD let's first try to prove the given quadrilateral to be parallel of RAM therefore angle 1 plus angle 2 is equal to angle 6 plus angle 8 so this is angle 8 isn't it and why is this since or you can write alternate interior angles we learned in lines and angles chapters right chapter alternate interior angle now angle 1 is equal to angle 2 and angle 6 is equal to angle 8 why given by sectors H F is by sector of angle CFG and gi is by sector of F GB so hence from these two we can say angle 2 is equal to angle 6 right angle 2 is equal to angle 6 but angle 2 and angle 6 happen to be alternate interior angle for which two lines H F and GI so hence we can say that H F is parallel to GI right again and alternate interior angles are equal and transversor is transversor is same FG angles are equal right so we can say that so H I H F is parallel to GI now let's go to this side now what I'm saying is next is similarly similarly what do we see angle 3 plus angle 4 right that is this angle is equal to angle 7 plus angle 5 that is this angle why again alternate interior angles so alternate interior angles is it again in this case we know that angle 3 is equal to angle 4 and angle 7 is equal to angle 5 why because of bisectors is it so hence I can say angle 3 is equal to angle 5 correct and because of this we can now conclude angle 3 and angle 5 are interior alternate interior angles so hence Fi is parallel to HG so using the first one we could prove that this line is parallel to this one and using the second we could prove that this line is parallel to this one that means what so let us write some identification equation number so you know so hence or rather we can do it like this is itself that therefore H F G Fi G or whichever name we had taken Fi GH right so Fi GH is a parallelogram we just proved it why because of this and this correct both the opposite pairs of sides are parallel so it's a parallelogram now now we will prove that one of them is 90 degrees so angle 1 plus angle 2 plus angle 3 plus angle 4 is how much 180 degrees linear pair so angle 1 plus 2 is one angle angle 3 plus 4 is another angle so it's 180 degrees and we know that angle 1 and angle 2 are same so 2 times angle 2 plus 2 times angle 3 is 180 degrees why because angle 1 is equal to angle 2 bisector is there and angle 3 is equal to angle 4 again bisector is there so that means angle 2 plus angle 3 is 180 degrees by 2 so this 2 is common to both isn't it so I can take that on the other side right right inside and divide so we get this is equal to 90 degrees so angle 2 plus 3 is 90 degree therefore angle H Fi is 90 degree and that is what we wanted so we have a parallelogram with 190 degree angle that means right so this is let's say 1 this is 2 so from and 1 and 2 we can see from 1 and 2 what do we conclude G or Fi G H was the name Fi G H Fi G H is a rectangle rectangle right so always remember a rectangle is nothing but a parallel graph with one of them one of the angles being 90 degree so hence using this property we could prove that EFGH is sorry not EFGH every time I am saying EFGH it's Fi G H is a Fi G H is a rectangle right I hope you understood this proof