 Hello and welcome to a problem-solving session on triangles The given question says that P is the midpoint of BC. So let us highlight point P This is P. It's midpoint of BC Q is the midpoint of AP right and If BQ when produced if BQ when produced meets AC at R. This is R Proof that RA is one-third of CA Okay, or RA by CA is one by three Okay, so that's the question right so how to go about this proof So first of all, let's do the customary activities, right? So given is P is the P is the midpoint midpoint of P is the midpoint of BC correct, therefore BP is equal to PC is equal to Same right or B. I can write BP is equal to PC. That's it, right? Similarly AQ is equal to QP This is given we have to prove to prove And I will be writing simplified version of the given proof AR by AC AR by AC AR by AC is equal to one upon three. So again, you can see AR by AC is equal to one upon three So there is a ratio of Two parts of the same segment, isn't it? So the moment ratio is there, you know what all theorems can be used So we have to do a construction, but here why because Mostly we are going to use the BPT, but in BPT we need a pair of parallel lines So what construction can I do? So if see AR has to be proven to be equal to one upon three That means this is one upon two Hmm and this is one is two one if you see I'm just writing crudely so one is two one one is two two We have to prove so by You know if let's say if I have this kind of a setup Let's say I'm doing construction PQ on a PS so PS parallel to QR drawn Right. So this will involve BP PC then there's a new point S also But then we'll get two parallel lines, right? So parallel lines and then we can write the BPT. So hence if you see proof What do we do in triangle in? Triangle which one so where do we see two parallel lines? So PS and BR are two parallel lines. So in triangle CBR CBR Right PS is parallel to BR So can't we say that PC by BP? Will be equal to CS by SR But PC and BP are same see same. They are same, right? No, so hence we can say This is equal to one Since PC by BP will be one Right. No, so hence what do we therefore we can conclude CS is equal to CR Right CS is equal to not CR SR my bad CS is equal to not CR SR This is SR, right? So just a correction guys it is SSR CS is equal to CS is equal to SR now again if you look at this triangle, which one APS in triangle APS See again, we have two Parallel sides parallel lines you have to be observant, right? So and if you see this is These are the two parallel sides or lines So in triangle APC QR is dividing the two sides of the triangle APS and PS happens to be the base Then we can say by BP By BP T what can I say? I can say a Q by QP a Q by QP is equal to AR by RS Now Q happens to be the midpoint So can I can I not say if you see a Q is equal to QP So this also will be equal to one because a Q by QP is one. So this implies AR is equal to RS and Previously we had CS is equal to SR or RS, right? So this is one This is two RS and SR are same. So therefore we can conclude from one and two from one and two What can we conclude we can say that AR is Equal to RS is equal to SC All three are equal. So this part equal to this part is equal to this part. So can we not say? Therefore we can conclude that AR AR upon AR upon AC is equal to AR Upon three AR because AC will be constituted or come, you know made by three AR. So hence it is one upon three Hence AR is equal to Or three AR is equal to AC Or AR is equal to One by three AC. This is what we needed to prove Right, this is what was the demand of the question C This one so we could Prove this so what was the underlying concept in this so again we use BPT But then we have to be observant in order to find two parallel sides and then since we are dealing with ratios, it's always Important or it's you know, it is quite quite evident that BPT will be applicable somewhere or the other so we have to just find the right configuration of the Geometry so right configuration. That means what do I mean by configuration? You have to find out those real, you know, those Required parallel sides. So the moment the parallel sides and the triangles are identified then applying BPT and then hence Coming to the conclusion will become easier. So learning is being observant and try to relate to the Requirement of the problem and the underlying concept