 Welcome friends We are going ahead with our series on gems of geometry and in this theorem. We are going to discuss One important property about pedal triangles and it is the third pedal triangle is similar to the original triangle That's the theorem and we are going to prove that before we You know get on to the proof. We would like to discuss Something about pedal triangle just a recap most of you all already would be knowing what pedal triangle is But just for a quick recap a pedal triangle is nothing But if we have a triangle and a point within the triangle So let's say this is a triangle and a point Let's say P and if you drop perpendicular on the three sides then The foot of the three perpendicular if you join them the triangle you obtain is called a Pedal triangle is it so that's what our learning has been so this is so ABC was a triangle and BEF is The pedal points What are pedal points foot of the perpendicular? Dropped from an internal point P on the three sides Those are pedal points and you join the pedal points you get a pedal triangle Now you can see that this point P Is lying in inside the pedal triangle itself So I can draw another pedal triangle with the same point P By dropping three perpendicular and keep on repeating this process So if you see this is this one is the first pedal triangle and This one is the second pedal triangle and if you keep on repeating the process Again, you drop a pendiculus. Let me do a little bit of zoom in Drop a pendiculus and join Yep, you'll keep on getting pedal triangle. So that's the Discussion here. So the theorem suggests that the Pedal triangle drawn for the third pedal triangle drawn will be similar to the original triangle. Okay, that's the theorem. Let's try and Prove this theorem. Okay, so how to prove this theorem. So all of us know that if these A1 B1 C1 are The pedal points for the first triangle, right? So you can see A1 here is A1 B1 C1 are the pedal points similarly A2 B2 C2 A2 B2 C2 are the pedal points for the second triangle. That means A2 B2 C2 is the second pedal triangle Similarly A3 B3 C3 is the third pedal triangle. So just to declutter the diagram we have removed The solid lines joining a A1 B B1 C1 and all so that you know you understand or the diagram looks a little neat But a A1 is 90 degrees. Okay, so keep it in mind Similarly B B1 is also 90 degree though It might not appear in the figure but B B1 if you join you'll get 90 degrees similarly C C1 and all these Point P to any pedal point will be a perpendicular. Keep this in mind Now let's see how to prove The two triangles are similar which two triangles are we anyways targeting triangle ABC is definitely one and Triangle A3 B3 C3 A3 sorry A3 subscript A3 D3 and C3 these are the three two triangles We are going to talk about and we are going to prove that these two are similar How do we know that two triangles are similar? We know that the criteria is Corresponding angles are equal than the two triangles are similar. Isn't it? That's the basic criteria for a triangle and so hence We need to prove that angle A is equal to angle A3 and angle B is equal to angle B3 and Angle C is equal to angle C3 if we somehow prove this Then we are done. Let's try and prove. So let's try in Look at points P a Let's look at P a B1 Okay, let's look at points he lies on the Circum circle of a B1 C1 I'm saying P Lies on the Circum circle Circum circle of triangle a B1 and C1 why is that because if you see AP Why is this I will just do for one and then similarly for all others it will follow so angle a C1 P is equal to angle a B1 P right and both are 90 degrees. So we can imagine that a P is the dia a P is the diameter A P is the diameter we can imagine that is it so hence P is lying on the circum center of a triangle a B1 C1 Fair enough. So So so hence using the property of a Circle, you know, so hence if you see angle P a B1 or let's see one a P Okay, so angle C1 a P this angle here will be equal to C1 B1 P this angle Right is equal to angle C1 B1 P and what is the logic logic is? a C1 P and B1 are Concyclic points so you can imagine like that. This is a circle and These are the points is it so hence This angle has to be equal to this angle right. This is C1. This is a this is B1 and this is P Right, so P was lying on the circum circle of a C1 B1 So hence angle subtended by arc P C1 at a will be equal to angle subtended by arc P C1 this arc at B1 right so by this logic C1 a P is equal to C1 B1 P now again if you see C1 B1 P Okay, now point B1 a 2 and C2 Again if by the same logic B1 a 2 and C2 right or let me write like this B1 Look carefully B1 a 2 C2 is the Or B1 a 2 C2 and P are again consicclic is it it let me write like that So I can write B1 a 2 P and C2 are Concyclic by the same logic By the same logic why because This angle is 90 degree This angle is 90 degree and hence this these four points are consicclic with P B1 as the P B1 as the diameter so hence I can further this by saying C1 C1 B1 P is equal to Is equal to a 2 B1 P and hence a 2 B1 P will be equal to a 2 C2 P a 2 C2 P by the same logic by the same logic which we used previously right now again C2 B3 and a 3 are consicclic. So P also lies on so I'm writing C2 B3 P and a 3 are Again consicclic Consicclic points why because again if you see This is 90 degrees. This is 90 degrees because of the pedal triangle, right? So P a 3 is perpendicular to B2 C2 P B3 is perpendicular to a 2 C2 So hence P B3 C2 a 3 Are consicclic points with PC 2 as diameter So hence again, I can further this and instead of a 2 C2 a 2 C2 P is equal to angle B3 C2 P because B3 lies on a 2 C2 and B3 C2 P Dear friends is equal to angle B3 a 3 P By this logic that angle subtended by an arc at two different points on the circle are equal right by this logic Okay, so we got what? Angle C1 a P is equal to angle B3 a 3 P, right? So let me highlight that so this angle We are getting it to be equal to this angle Okay Now if you apply the same logic dear friends, you will be able to let's say this one was x Then you will be able to prove that this angle is x So hence I am rewriting this for your understanding. So let's say I am writing angle P a B1 P a B1 is equal to angle P C C1 B1 Okay, where P B1 is the card. So if you consider P B1 as the card so P a B1 will be equal to P C1 B1 now P C1 B1 is equal to P C1 a 2 Right because a 2 lies on B1 C1 right P C1 a 2 now PC1 a 2 guys will lead to angle P B2 P B2 a 2 Right P B2 a 2 why because P a 2 C1 B2 are again consicclic by the logic which we have been using so far again now if P B2 a 2 is equal to angle P B2 C3 and Now if P B2 C3 will be equal to angle P a 3 C3 Okay, why because P C3 B2 a 3 are Ciclic okay, so always remember you have to draw these kind of triangles if you're not getting on these these kind of diagrams If you're not getting it straight. So hence if you do this multiple number of times, you know One by one you can equate angle subtend by the same chord or arc on two different points of the circle You will see that you are getting the equality and hence eventually what do I get from these two relations? Let's say this is one and this is two so hence I can say from from one and two angle P a a C1 is Equal to angle P a 3 B3. This is the first relationship and Second relationship is P a B1 is equal to angle P a 3 C3 Now guys, let's add both of them. So you'll get angle C1 a P. Let me write like that plus angle P a B1 is Equal to angle B3 a 3 P plus angle P a 3 P a 3 C 3 now. Let's see what all these angles are so C1 a P Plus B1 a P is nothing but angle a itself. So hence I can write angle a and B3 a 3 P B 3 here is B3 if you see this is B3 B3 a 3 P plus C3 a 3 P will be nothing but B3 a 3 C3 Will be equal to nothing but angle B3 a 3 C3 this is what we have to prove right so hence we get in these two triangles angle a is equal to angle a 3 similarly, it will not be tough to prove that Angle B will be equal to angle B3 and angle C is equal to angle C3 hence by a a a similarity criteria similarity criteria triangle a b c is Similar to triangle a 3 B 3 C3 that means the third pedal triangle is similar to the original triangle. That's what the theorem was suggesting