 Hello friends welcome again to another session on similar triangles and today we are going to learn one of the very important theorems in all geometry and it is called as basic proportionality theorem, which was which is also named after Thales Thales, and he is pre-Sopratic philosopher, so he existed before Socrates existed in Greece and he is also considered one of one of the seven sages Seven sages, seven sages in Greece Okay, so this is a small introduction about this guy So this this was a scholar and a philosopher who gave a lot of mathematical theorems and In and especially in geometry Now let us understand what the basic proportionality theorem is. It is also called as intercept theorem. Okay, so the the Statement of this theorem goes like this and the statement is I'm now writing the theorem statement and it says if A line if a line is drawn if a line is drawn parallel parallel to one side one side of a triangle of a triangle intersecting Intersecting the other two sides Okay, other two sides right then then It divides It divides divides the two sides the two sides In the same ratio In the same Ratio, so this is what is the statement for Tell you sir or basic proportionality theorem now. Let us understand this theorem What does it say? So first of all it talks about a triangle. So let us have a triangle first So let us say I have a triangle like this This is a triangle. So let me draw a triangle This is a triangle. Okay, and now it says that if a line is drawn parallel to the two sides So I'm drawing a line parallel to these two sides and the other criteria is it must also intersect the two sides Sorry parallel to one of the sides. So hence, let us say this triangle is a bc a bc and I'm drawing de parallel to parallel to bc in this case then Then and then yes intersection is also important, right? So de must intersect the two sides that is ab and ec why because other line parallel to bc could be simply this Then there is no question of ratio only right another line could be like that parallel to bc But the criteria is the line must be intersecting the two sides. So now de is clearly intersecting the two sides of the triangle then so According to Thales theorem de is parallel to bc Okay, so hence ad by db will be equal to ae by ec This is what Thales theorem say and it need not be only that de is parallel to bc. Let us say I can draw another line parallel to Let us say ab. So let us say this is the line parallel to ab. Let us say this is pq pq then then also so let us say if pq is parallel to ab Then cp by pa is equal to cq by qb Okay, another example, let us say this is the line parallel to ac and let us say this is x y Two points are x y. So hence again bx upon xc is equal to by Upon ya okay, so Bx upon xc is equal to By upon ya. So there though any so there could be three possible Parallel lines to the triangle right now intersecting the other two sides So in any which way any which case this is what you have to You uh, this is what is the result. So cp upon pa is equal to Uh cq upon qb ad upon db Is equal to a upon ec And bx upon xc is equal to b by upon ya, right? So this is what is the description or let's say all about what Thales theorem is okay Now, let us see the proof of thales theorem. How do we prove thales theorem, right? This since this is a theorem there must be a proof for it. So let us see how So let me redraw a triangle and Yeah, so this is a triangle and let us say This line is parallel to One of the sides, right? This is a b and c Okay, let me Now what I have now given what is given given is Let us say this is d and e. So given is d e parallel to dc, okay, and what is to be proved to prove To prove ad by db Is equal to a e by ac Now to prove this we have to use some construction. So what is the construction? First of all drop a perpendicular from E on to ab let us say this point is f And drop another perpendicular on That's a ac from Let us say this is g So construction What is the construction? dg perpendicular to ac and E f perpendicular to ab Okay, and another construction is join Be Be and dc joined Okay, so I am Right, this is the construction now. We are going to use the concept of area To prove this theorem. So let us now write what is the ratio of area of Triangle ad e divided by area of triangle db e Okay, what is area of triangle adb is nothing but half into base Now if I take ad as the base Then fe is the height right because E f is perpendicular to ab we have just drawn construction, right And similarly dbe if you see if you take db as the base then again E is the third vertex if you drop a perpendicular on to db. So e f is the perpendicular. So you can again write fe Right. So if you see this this one here This one is perpendicular to both db as well as ad isn't it So half into base into height. Okay. So this fe fe goes half half goes so hence we get ad by db Now if you see we have just got the elitist of the thing to be proved ad by db Somehow if I get a e by ec and you know somehow we can prove that both are same then You're done Now similarly I can take What do what can I say I can say? area of same ade divided by area of triangle cde Cde look carefully cde i'm going to highlight those this area cde this is the area And the top one ade right if you see What is it again half into base now this time around I will take base as a e Why because I need somehow a right half into a e into what is the height height is dg Isn't it height is dg and divided by half into base of cde this highlighted triangle if you see base is c e And height is again dg why because d is the vertex the third vertex of cde And from d you are dropping up a pendicular on to ce which is nothing but dg So again this dg this dg goes Half half goes so hence if you see it is nothing but a e upon ec or ce Correct. Let me name this equation one Equation two Now we are definitely getting what we need Now we are getting definitely what we needed to prove Correct, but then for that if we somehow prove these to be equal Then our job is done Now if you see In the LHS of both one and two the numerator is same and which is d ade ade That means if I somehow prove the denominator is also to be same Then my job is done and hence the RHS must must also be equal then Let us see whether triangle area of triangle dbe. What is area of triangle dbe? So let me Reduce the size a bit and yes, so so that we can fit everything in the same frame Now if you see we let us consider area of triangle dbe Area of triangle dbe is nothing but half into d e Half into d e d is the let's say base and you can drop a perpendicular from B on to d e. Let us say this height is h So half into d into h Now area of triangle c d e if you say Is also half into d e again. I am taking the same d e And if you see c d e also is between the same parallels. So hence again here also This is the perpendicular. So this height will also be h because Distance between two parallel lines is always same So into h that means from this argument I can say What can I say? I can say area of triangle dbe Is equal to area of triangle c d e. Isn't it and hence Let us say this is three So clearly from one two and three from one two and Three we can say ad by db is equal to a e by Easy right hence proved Once again, how did it happen? We found out the ratio of the two triangles. We got this one We got this one and we had to eventually equate these two. Yep for that The LHS of both must be same We saw that the numerator is same We just proved that the denominator is also same And the denominator is nothing but area of two triangles Which are same because the base is same and the height is also same Correct hence the LHS part LHS part is same So if LHS part is same of two equations Then RHS part has to be same and that's why we arrived at this result Okay, so this is called basic proportionality theorem right once again But it's basically proportionality theorem or Thales theorem or intercept theorem. It is nothing but Uh, if a line joins or or line is parallel to one side of the triangle then it divides the other two sides into equal ratios