 Hi friends welcome to another problem-solving session on triangles and In this question, it's given that any point X inside triangle DEF is joined to its vertices So X is a point inside the triangle and it is joined to all the three vertices PENF from a point P in DX PQ is drawn parallel to DE so first information is PQ is drawn parallel to DE and Meeting X E at Q right so this point is Q and from Q QR is drawn parallel to EF So this is these are the two parallel lines Such that this point is R. You have to prove that P R is also parallel to DF Okay, so this is the question now after reading the question carefully We'll have to understand what method is to be adopted Now again as we have been seeing in all these questions, it's Certainly a case of a triangle and there are lots of parallel lines Parallel lines within Triangles and the lines are one of the parallel lines is the side of a triangle and then We have to you know conclude that another line is parallel to the third side So the moment these criteria are there. So the first thing which comes in mind is B P T or Thales theorem Thales theorem Okay now So how to apply Thales theorem here and what is Thales theorem by the way, so you know if there's a triangle and There is a line parallel to one of the sides then Let's say a B C D E So, you know by Thales theorem a B By DB is equal to a E by EC and the converse of Thales theorem is also true Converse of Thales theorem is that if this is a certain that this is the ratio then the line which is you know Cutting the two sides at point D and E will be parallel to BC the so hence if this is This goes to here. So this also comes back to here Okay This is what We have learned in in in case of Thales theorem and Converse of Thales now So consider triangle X D E right this is a triangle where you see there are two lines parallel What is what are the parallel lines P Q is parallel to? D E. So what can we write about it? So we can write X P upon PD XP upon PD will be equal to X Q upon X Q upon Q E Let this be relation number one and why is this this is by by basic proportionality theorem Correct now again in triangle in triangle X E If again, so we are picking up that triangle Where I can go ahead with this ratio. So again if you see X Q again in this triangle What will be this? X Q upon Q E will be equal to X R upon RF This is equation two. Let's say so you can now write from one and two From one and two. What do we get we get? XP by PD is equal to X Q by Q E. So XP by PD is equal to X Q by Q E by Q E is equal to X R by RF That means I can equate these two as well that means XP upon PD is equal to X R upon RF Isn't it? Right now if you see XP upon PD is equal to X R by RF What does it mean? It means that this side by this side is equal to this by this so hence by Converse of Converse of BPT we get PR is parallel to FD Isn't it? This is what you needed to prove by Converse of BPT. What is Converse of BPT? We just learned that if in a triangle If this is a triangle and there's a line like this and let's say A, B, C and Let's say D and E and it is given that AD by DB is equal to AE by EC then this results This means that DE is parallel to BC. This was Converse of Thales theorem Correct. That's what we used here and hence we got PR to be parallel to FD I hope you understood the solution. So try some more problems similar to these types from a standard textbook