 Hi, so in this question it's given that AB is parallel to EF and EF is parallel to DC and That means this particular diagram ABCD is a trapezium. So it's a trapezium now You have to prove that AE upon AE upon ED is equal to BF upon FC Now the moment we see such kind of a question and with the diagram if you see There are parallel lines and you are asked to prove two ratios to be equal now if you really Notice this is basically ratios and parallel lines. Where have we Encountered such such kind of case definitely in a triangle and we have learned something called a basic proportionality theorem Is it a basic Proportionality theorem in a in a triangle and what does it say? So if you have a triangle like that and you have a line parallel to one of the sides, let's say, this is a BC and This is D and E So, you know if DE is parallel to BC then AD upon DB AD upon DB is equal to AE upon EC. Isn't it? This is what we have learned about a Triangle or basically now the basic proportionality theorem now Though we have ratios and parallel lines here, but You know, we don't see any triangle here. So what could be the possibility? So hence basically we need to carve out a triangle over here So either there are two ways to solve this problem. Either you produce DA and CB and let them meet at a point. Let's say P Yeah P this is one way of doing it or I would prefer the other and So basically let me complete the first part. So hence if you come if you join them So P is the point of intersection of DA and CB intersect extended So now if you see there are two triangles. There are triangle P a P e P e f and triangle P D C So in these two triangles applying BPT, you can do it do it. Otherwise, what I will Try and do is I will join these points AC and will apply the BPT here. So let me join them. So I'm joining these two points Okay, so I'm joining AC. This is a construction. I did Now again, if you see there are two triangles we obtain and things will become simpler and let this point be G Okay, the point of intersection of EF and AC is G. Now in triangle a D C Okay, let's consider triangle ADC Clearly EG is Parallel to DC. Why because it's given that EF is parallel to DC Right then Then by Thaley's theorem by Thaley's theorem, what can I say? I can say a E upon ED is equal to a G upon GC A G upon GC let that let this relation be one now now Let us now write it here. So in triangle ABC if you see this EG again is parallel to the third side AB So hence we can say GF is parallel to AB Isn't it? This means by BPT again by Thaley's theorem or BPT whichever way you want to write Thaley's theorem again What can we say we can say? CG upon GA is equal to CF upon FB and applying inverted on to it or you Take the reciprocal. It would be same. So GA upon GC is equal to CF upon FB I'm sorry, not CF upon FB FB upon CF, right? So this will be FB. So I took the reciprocal FB upon CF or FC whichever way now this is Relation number two So from one and two guys if you see here also on the right hand side, it is AG and GC AG by GC Here also it is AG by GC. So hence from one and two from from one and two you can say a E by ED is equal to a G by GC AG by GC is equal to FB by CF hence hence a E by ED is equal to FB by CF, right? And this is what we have to prove Isn't it? So We have to prove a E by ED is BF or FB by FC So this is what we needed to prove and we ended up getting the same thing It's only thing is the only thing is this is FB is written as BF over here All right, so hence if you can match this one and this one are Same, right? So in this question, what did we learn? We learned that though it was not very obvious That we have to apply basic proportionality theorem but the very given fact that there are parallel lines and The expected result was to prove a ratio. It was prompting that we have to use DPT somehow and we tried that and it helped and the problem was solved