 Hi and how are you all today? I am Priyanka and let us discuss this question. It says, in figure 9.33, this is the figure, 9.33 which we need to refer in this question, A, B, C and D, D, E are two equilateral triangles. This triangle and this triangle is an equilateral triangle such that D is the midpoint of B, C. D point is the midpoint of B, C. If A, E, A, E intersects B, C and F show that these are the six subparts of the question which we need to show and we will be proceeding it one by one. Let us start with our solution. Here for the whole question we are given that triangle E, B, C and triangle B, G, E are two equilateral triangles, right? And also G is the midpoint of B, C. So that means B, G is equal to D, C. We need to prove six things. So let us start with the first one or let us avoid writing this section. One of the things that we need to do over here is we need to join A, D and E, C. So this will be a part of our construction. That is to join A, D and E, C also draw E, M or pendiculato. Now if I just have this figure again, this is the figure. What we need to do over here is we need to draw E, M or pendiculato B, G. Then we need to join A, D and E, C. Now let us start with our first parts prove. Here, what we need to prove is that area of triangle B, D, E, B, D, E is equal to one fourth the area of A, D, C. That is this mid-triangle. Let us divide our page into two columns. Let this be the left-hand side and this be the right-hand side. First of all, in the left-hand side, we will be finding out area of, let us write it in short form only. That is area of triangle B, D, E. We know that the formula for finding out the area of a mid-triangle triangle is root 3 by 4 side square. We can take any of the side here. We can write root 3 by 4 and let us take this side as B, D, square. Similarly, here we will be finding out area of A, B, C. Since in the L, H, S, we have one fourth. Let us find out one fourth area of A, B, C. That will be one by four. Again, using the formula for an equilateral triangle, that is root 3 by 4 side square. Here also we can take any of the side. So let us take side as B, C. So here we even have B, C, square straight away. Now, further, we can solve it like root 3 by 4, now here B, D can be written as B, C by 2 as it is half of B, C. G is the midpoint. So we can have it like this. Then we have root 3 by 4, B, C, square by 4, getting multiplied. And further, we have root 3 by 16, B, C, square as our L, H, S. Whereas talking about our R, H, S, we have root 3 by 16, B, C, square straight away. Since L, H, S is equal to R, H, S, so we can say that we have proved the first part. We have proved that area of B, G, E is equal to one fourth area of A, B, C. Right, so this completes our first part. Now, let us proceed on with the second part. The second part which we need to prove is that area of B, D, E is equal to half the area of B, A, E. Let us start with our proof. Now, for this, we should be having a diagram. Now, if you notice that an equilateral triangle has all the angles equal to 60 degrees. So all the angles are equal to 60 degrees. Now, we have C, V, A, that is C, V, A. This angle equals to 60 degrees. And also, we have angle B, C, A equal to 60 degrees. Again, we have C, V, E, C, V, E, this angle equal to 60 degrees and this angle also equal to 60 degrees. So if you know till they are forming, alternate angles. So we can just write down that angle C, V, E is equal to 60 degrees because triangle V, D, E is an equilateral triangle and similarly, angle V, C, A is equal to 60 degrees because triangle A, D, C is an equilateral triangle. So we know that it is said to be parallel when two lines are intersected by a transversal. Now here, the two lines we are talking about is A, C, and V, E. When these two lines are intersected by a transversal B, C, such that alternate in till angles are equal, then the lines are parallel to each other. So since these two are parallel and also we notice triangle B, E, C we can say that area of triangle B, C, E, B, C, E or B, E, C is equal to area of triangle B, A, C. B, they both are on the, are equal because of the simple reason because they are having a common base that is V, E and they are between the same parallel lines. So we can say that area of those triangles which are having a common base and are between the same parallel lines are equal. Further, we can say that area of B, G, E is equal to half the area of B, E, C because B, G, E is half the area of B, E, C. Because E, G is the median and the median divides the triangle into two equal half. So B, E, D and G, E, C are two equal half of the thick triangle. Let us have a small figure that will help you in getting it correct. This is B, this is C, this is E. E, D is the median because D is the midpoint of B, C. So we can write that B, D, E is half of this thick triangle that is B, E, C, right? Further, above we have proved that area of B, E, C is equal to B, A, E. So we can just easily substitute it and write down that it is equal to half of area of B, A, E and let this be the first equation and this be the second equation and here we have proved the required second part. Now the third part which we need to prove is area of A, B, C is twice the area of B, E, C. Now, we know that area of B, C, E is equal to half the area of B, E, C. This is proved in part two, that is the above part. Also we know that area of B, T, E is again equal to one by four area of A, B, C. That is proved in part one. So by these two we can say by comparing third and fourth equation we have half of area of B, E, C is equal to one by four area of A, B, C because at the left side of these two equations are same to each other so therefore the right hand side will be equal to each other. Now if you take this two from the denominator to the numerator we are left with area of B, E, C is two by four area of A, B, C which on simplifying will give us one by two or this two will again be taken to the left hand side so we can say that or twice area B, E, C is equal to area of A, B, C or reversing it we have area of A, B, C is equal to two times the area of B, E, C. This is the proof of the third part and we have proved it. Now let us proceed on with the fourth part which we need to prove over here is area of B, F, E is equal to area of A, F. We proved that these two lines are parallel to each other similarly we can prove that this line is parallel to this line because again alternate angles are equal to each other. We can say that straight away G is parallel to VAB because angle A, B, C is equal to angle E, T, B is equal to 60 degrees and also they are alternate angles between these parallel lines. So the triangles, triangle D, D, E that is the area of this triangle B, D, E is equal to area of A, E, T by the use of theorem 9.2 which says that the triangles having common base and are between the same parallel lines are equal to each other the area of equal to each other has a common base is D, E and the parallel lines are D, E is parallel to AB. So these both triangles are between are having the same base and are between the same parallel lines hence while theorem 9.2 we can say that the area of are equal. Now on subtracting area of GSE from both the sides we have area of PGE is equal to area of A, E, G as they have proved above and if we subtract area of GSE from both the sides we are left with area of PFE is equal to area of A, F, T which we were supposed to prove. So here we have proved the fourth part also. Proceeding on to the fifth part very quickly. Here we need to prove area of PFE is equal to twice the area of FEG. Let us start with our proof once again. Now if you notice that G is the midpoint of BC and triangle ABC is an equilateral triangle so we can see that therefore AD is perpendicular to BC and also we know that by construction we have EM this is the point M is perpendicular to BD. Now let us first find out the values of AB and EM and let the length of AB be equal to X. Now by Pythagoras theorem we can say that AD is the square root of BC. If we say AD is perpendicular to BC then AD is AC square or AB square minus BD square. Now AB we have taken to be as X centimeter so it will be X square minus X by 2 the whole square because BD is half of AD as we have proved above. On simplifying it we have that the value of AD is root 3 by 2 X. Similarly EM is equal to X by 2 the whole square minus X by 4 the whole square and it is X by 4. So here on solving we have X square by 4 minus X square by 16 which is equal to 4 X square minus X square by 16 3 X square by 16 then the root that is equal to root 3 X by 4. So on comparing 1 and 2 we have that AD is equal to 1 by 2 of EM because AD is root 3 by 2 X and it is equal to 1 by 2 or it is twice EM 3 by 4 X. So on tankling it will be equal only. Further in triangle we know that area of triangle AFT is equal to area of triangle FED. If you find out their area it will be half into base that is FG into height that is AD whereas it is also equal to half into base that is FG and height that is EM. We will take the perpendicular heights only. Now if we substitute what we got from above we have 1 by 2 into FG into AD is equal to 1 by 2 into FG into EM can be written as NAHER. We found out that AD is equal to twice EM that was 1 by 2 EM because when we multiply AD by 1 by 2 then only we will be able to get EM. That is why it is NAHER. We were comparing them so if you notice it should be half into FG into NAHER. EM can be written as 1 by 2 AD and if you see the only difference is that this is 1 by 2 AFT triangle. The area is twice this triangle because this is half of it. So twice area of FED is equal to area of AFT. But we also have that area of BFE is equal to area of AFT in part 4 we proved it. So since right-hand sides are equal we can say that the left-hand sides are also equal and we can say that therefore area of BFE as is this is equal to twice area of FED. And we have proved the required question. This completes the fixed part and the last and final part that is the fixed part says we need to prove over here is area of FED is equal to 1 by 8 area of AFT and with the help of the other proofs that we have done we can say that we have area of PDE equal to 1 by 4 area of ABC. This was in part 1 that we proved. Also we can write that area of PDE can be split up and written as area of BFE plus area of FED is equal to 1 by 4 area of ABC can be written as twice area of AGC because this is half of this whole triangle. Further we can write that twice area of FED because this triangle is twice this triangle plus area of FED is equal to half that will be on C. Simplifying area of ABC can be written as area of AFC minus area of AFG. You can have a picture to simultaneously see the required happening further. We have twice the area of FED on adding these two parts. That is these two is equal to half area of AFC. We have multiplied half with this, multiplied by half into area of AFG can be written as twice area of FED which we have taken from the above part or four times because on cancelling out we get 1 and when we will be taking this from license side to license side become four times area of FED is equal to 1 by 2 area of AFC and if you take this four from license side to license side we have area of FED is equal to 1 by 8 area of AFC which was needed to be proved and with this proof we have completed the given question. This was a big question but with the help of the above proofs you can easily prove out the next proofs also. Take care.