 Hello my friends. So welcoming again to a new session on problem solving I know triangles is a big, you know Nightmare for so many people but the only way to deal with it is solve as many problems as possible And you'll see all these troubles melting down Now in this given question, it's been given that ABCD is a parallelogram So let us read the question carefully ABCD is a parallelogram the moment The word parallelogram comes, you know opposite sides are Equal and parallel opposite sides are equal and Parallel isn't it? So this is what we have known about the properties of a parallelogram P is a point on site BC. So this point is P deep and DP when produced Meet AB produced at L nothing to worry about it or very easy to understand And now what do you need to prove is that DP upon DC? That means this DP upon DP upon sorry DP upon PL So, okay, and should be equal to DC DC upon BL Okay, this is what you have to prove and DC is nothing but if you see AB so In a way while Reading the question itself, it is pretty clear that it is going to be true Okay, so hence direction is very clear and what is the first thing which comes to our mind the moment we see such a problem So if you see there are parallel lines one is parallel lines, right? And that too, there's a triangle inside a triangle So triangle these two combination and then we are dealing with ratios as well So hence it hints upon using basic proportionality theorem basic proportionality Proportionality theorem and What was that dear friends nothing, but if you have a triangle you have a parallel line. So a B C D E So by BP T if D is parallel to BC then AD upon DB is equal to AE upon EC no problems in that, right? Now here if you see We have similar type of a triangle here and that triangle is so now we will say in Triangle A LD If you see BP is parallel to AD Isn't it why because BC is parallel to AD and BC and parallel to AD because they're there are parallel ram So this side is parallel to the side, right? therefore by BP T in this triangle we can say We can say P L P L by DP Right, so this side by this side is equal to is equal to A PL by PL by DP or sorry D PL PL by DP is equal to BL by AB Is equal to BL by AB so I'm let me just highlight this triangle what I'm trying to say is this triangle we are considering and in this triangle there is a Parallel line. So hence it is similar to this case, isn't it? this case right so hence I Could write that so if you invert it that is take the reciprocal of both sides you will get BP by PL is equal to AB by BL Right and this process is called invert endo We know this what is in the endo a by B is equal to C by D. This means B by a is equal to D by C Okay, so this is what it is DP by PL is equal to AB by BL Okay, now we'll again use the property of the triangle of a parallelogram. Sorry now in this case What was to be found out? Let us just keep the target closer to us so that you know, we always can so let me copy and Let me copy this target So that yeah, so hence now the target is here for us to see okay, so hence You see for the first problem the LHS DP by First in the first problem the LHS The LHS here. Yeah, so hence if you see the for the first problem DP by PL is already here and on the right-hand side BL is there in the denominator as well here But only difference is AB and DC. So we also know that AB is equal to DC Since if they are sides of opposite sides of a parallelogram, hence clearly we can say DP by PL is equal to DC upon BL so hence proved first one is proved Why proved because AB is equal to DC. Why AB is equal to DC because AB is opposite Side to DC in a parallelogram. So the first one is done What about the second one? So if you see second one We have DL by DP so first understand from the diagram itself Where are they lying? DL by DP. So let us say DL. Where is DL guys? So this entire thing is DL and They're saying DL upon this thing is equal to AL is this whole upon DC that is this one, which is nothing but AB right, so it looks like corollary to the corollary to Bpt. So hence again, you can say in triangle APD or rather you can start from here itself You see So what do we know? We can start from see we want DP in the denominator So we have one relation here where a DP is in the denominator. So let's start from here. So hence. I'm saying PL upon DP is equal to BL upon AB Now if you keep the target in mind AB is any ways equal to DC, isn't it? Right, AB is equal to any way DC. So we are getting the denominators to be equal We have to just see how to get DL on the top now add one to both sides. So DL. What is DL? DL is DP, which is the denominator plus PL Right, this PL is added to DP to get DL. So hence I have to add that so how to get that So I'm writing adding adding one to both sides Now this is an important step in many such question adding one because that will give you different different ratios So hence, what can I do? I can say PL plus DP. Okay before that PL plus one By DP is equal to BL by AB plus one Isn't it? Now let's take the denominators common. So if I take DP as in the denominator, I'll get PL plus DP Here it is BL plus AB upon AB Now clearly PL plus DP. This PL plus DP you check it is nothing but DL. So hence I can write this as DL DL upon DP and what is BL plus AB guys? BL plus AB from the figure if you see BL BL Plus AB is AL. So hence I can write this as AL by AB Now keeping the target in mind C DL, DP, AL Three things we have got DL, DP and AL only DC is left, but we know that AB is equal to AB is anybody equal to DC. So hence this is nothing but DL upon DP is Equal to AL upon instead of AB. I can write DC. So hence we got the result Okay, so both results were Done so learning learning is first of all the theorem the basic theorem underlying concept is basic proportionality theorem And then we also need to observe the patterns the ratios and we also must be ready Or must be geared up to work with ratios, right? So working with ratio also is an important You know in ingredient for solving these problems. So hence, please Know how to work with ratios and the fourth thing which you must know is Componento, componento Componento, dividendo Dividendo, dividendo, invertendo And alter, nendo, alter Nendo, we will talk about these ratios You know those operations on ratios in a separate sessions, but you must be aware of all these