 Hello and welcome to another problem-solving session in geometry and we are solving problems related to triangles and so far we have seen the basic proportionality theorem and internal angle bisector theorem of a triangle and external angle of a triangle Bisector theorem, right? So these are the theorems which we have studied so far Now here is a question which says that abcv is a quadrilateral in which ab is equal to ab is equal to It's not cd actually just a correction. It's Let me just correct it angle Bac and cad Intersect decides BC and cd at the points ENF respectively prove that EF is parallel to Bd so in such questions, let's say diagram is not given. So let's first draw a diagram. So here is D diagram, okay, so this is let's say looks like a parallelogram, but it's okay, right? So hence we are saying this is a This is B. This is C This is C and this is D Okay, this is D. Now what? We have to draw these angles bisector of Bac. So first of all, let's join AC So I'm joining AC. So this is AC and we have to draw a bisector of bisector of Bac. So let's say this is the Bisector and similarly of Cad. So let's say this is the Another bisector. Okay, so let's what's the name. So the bisector of Bac That means so we have to put a name or at the points ENF respectively. It says so this is point E and This is point F and they're asking us to prove that EF is parallel to Cd Bd, sorry right, so This is what you have to prove. Okay, so let's try and prove so before that it's And a good practice to understand the approach first So we have to prove what C, EF parallel to Bd, isn't it? Now, what have we learned about parallel Sides and parallel elements within a triangle. We have learned that what is that that in basic proportionality theorem if you see if This ratio FC by Fd is equal to CE by EB somehow if you prove Then by the converse of basic proportionality theorem EF will be parallel to Bd, right? So this gives us some sense and now since we are also dealing with internal angle bisector theorem So we know that DF by FC will be equal or be equal to AD upon AC it has to be then only Right because ASAF happens to be the bisector in Triangle ADC correct. So hence this ratio should also be same I will write anyways, don't worry But what we are trying to do is we're trying to assess whether we reach out to that given Demand of the question by so we are just going in the reverse direction and see we end up at the given conditions, right? So in the other case This CE by EB must be equal to this AC by AB Correct because AE is the angle bisector So looks like we can approach like that because AD is equal to AB given so it will help us. So now Try to do it formally. So what is given? AF bisects angle BAC right and AE bisects angle BAC Okay, this is given also what is given AD is equal to AB These three things are given Correct now we have to prove To prove what we have to prove that EF is parallel to BD Okay, we have done the Reverse steps and we know that somehow we can do that using internal angle bisector theorem and converts of BPD. Okay, so let's prove it We don't require any construction in this case now You can say in triangle See ADC ADC since AF bisects Angle a B or let's say DAC not B and all angle DAC DAC therefore we can say AD upon AC is equal to DF upon FC and what is the logic? logic is internal angle bisector of Triangle theorem Right, we studied this in previous Sessions right so AD by AC is equal to Let it be one map Come to this side. Okay, so now we are saying in triangle BAC now BAC triangle BAC B a upon AC Is equal to be upon EC B a upon AC is equal to BE upon EC same reason internal angle bisector theorem since We know that since AE is internal angle bisector angle bisector of Angle BAC because of that right So this is fine. This is two now from And also we know that AB is equal to AD given Let it be three And it's given Given condition isn't it? So now from one two and three if you look at it from one two and Three what can I say? I can say Let us first have a look on one so AD by AC AD by AC so can't I replace AD by AC this can be written as AB by AC Isn't it DF by FC Right, AB by AC why because AD is equal to since AD is equal to AB Now if you see this let us call it four So instead of writing from one two and three let's write from four from two and four Two and four. What do I get see LHS is same of two and four So hence RHS will also be same. So DF by FC is equal to BE by EC and by reciprocating you can say FC by DF Is equal to EC By BE I just reciprocated just for a better view of it. Nothing nothing more the proof is concluded here itself as just for you know better view So FC by DF is nothing but FC by DF is equal to EC by BE. Is it it? This is what it is. So by converse of by converse of Basic proportionality theorem. I'm writing in short basic proportionality theorem or Thaley's theorem You can prove that EF is parallel to B B Right hence proved Hence proved and why will they be parallel because the converse of Basic proportionality theorem suggests that if a line in this case FE Divides the two sides of the triangle that is CD and CB in equal ratios Okay, so equal ratio means if this is X. This is why This is also X. This is why so X is to Y then FE will be parallel to the third side DB So these two sides will be parallel. That's what we used it and hence proved this particular question, okay