 Hello friends, so in this session we are going to solve another CBC problem This is CBC 2010 mathematics paper problem, and it is related to geometry So the problem says in the given figure if PQ is parallel to BC, so This point is C this point is C So hence in a given figure PQ is parallel to BC and PR is parallel to CD So this PR is parallel to CD and PQ is parallel to BC They prove that AR by AD so this this AR upon AD is equal to AQ upon AB So those who have dealt with the triangles and similar triangles and Theorems related to that they will you know it will strike to them in the very first instance of Looking at this problem that this problem requires the use of Basic proportionality theorem. So what is basic proportionality theorem? Just a quick recap. So basic Proportionality Proportionality Theorem which is also called the Thales theorem Thales theorem Thales theorem, what is this? It's nothing, but if there is a triangle if There is a triangle ABC ABC and There's a line DC DE sorry DE and DE is parallel to the third side BC then then we know that AD upon DB is equal to AE upon EC This is basic proportionality theorem. So we'll be using this theorem to prove this problem. Okay now so what's given so Given you should write first given as whatever is given. So if you can see PQ is parallel to BC PQ is parallel to BC. This is one and then PR is parallel to CD Okay, now to prove You should be writing These steps so to prove first is a R upon AD is equal to a Q On a B. Let's try and prove this first. So how to prove so proof. Let's Try to find the proof So if you see in triangle a BC But it's given it's given that PQ is parallel to BC Given here PQ is parallel to BC. So if that is so then you can write by using by using Thales theorem or basic proportionality theorem in the first instance, please try to write the theorem's name in full basic proportionality Proportionality theorem and then you can always put the acronym Along with it so that you can use it later on. So by using Basic proportionality theorem, we know that what do we know a Q upon Qb will be equal to a P upon PC, isn't it? Similarly, you can say similarly Similarly In triangle ADC ADC AD Sorry PR is parallel to CD. It's given. Where is given here? It is given PR is parallel to CD then using Using BPT you can write now using BPT you can say a P upon PC Is equal to a R upon RD Isn't it? So this let this be equation number one and this be relation number two and now We'll proceed further. So you can say now using using one and two We can say If you see AP by PC AP by PC appears in both the relation so you can write a Q upon QB is Equal to a R upon RD AQ upon QB is equal to AR upon RD I have to prove AR by AD actually. So you have to prove AR by AD. So what I can do is The same relation can be written as if you inverse it so invert into you'll use QB by a Q is Is equal to RD by Why did I do do this? You'll get to understand in a moment. Now what you do is add one to both sides So QB plus a Q plus one is Equal to RD upon AR plus one and this will lead to QB plus QB plus a Q Divide by a Q Is equal to RD plus a R Divide by AR now, let's have a glimpse of the picture so that you can Understand it better. So if you see from the figure QB plus a Q QB plus a Q will be nothing but AB so I can write it here AB and AQ stays in the denominator RD plus AR AR plus RD is Nothing but AD and then this is AD upon AR AD upon AR and then again you can invert them to say a Q upon AB is equal to AR upon AD AR upon AD is equal to a Q by AB. So this was what was required as first Yeah, so hence proved The first part is proved Okay, what about the second one second one says QB upon a Q so QB QB upon a Q so if you see you just need to In fact, you had already proved prove that why because QB upon a Q you can say a Q upon QB Here it is already proven right so if you see QB upon a Q is equal to RD upon AR so this has already been proved Here, yeah, so you can hear here itself. You would have mentioned that This is proof for Proof for part two Okay, so these two relations were to be established and then you have established it so what's the learning so basically the moment we see triangles and parallel lines and ratios around sides are You know involved then, you know, there's a hint that you can use basic proportionality theorem or Thiele's theorem to prove the required Relations. Thanks for watching this video