 Hello and welcome to the session. In this session we discuss the following question which says proof that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sites. Using the above result, proof the following. In a triangle ABC, XY is parallel to BC and it divides triangle ABC into two parts of equal area. Proof that BX upon AB is equal to root 2 minus 1 upon root 2. First let's recall what is AA similarity criterion. According to this we have that if in two triangles two angles of one triangle are respectively equal to the two angles of the other triangle then the two triangles are similar. This is the key idea that we use for this question. Now let's move on to the solution. Let's see what is given to us in the question. Then we have triangle ABC and triangle DEF such that triangle ABC is similar to triangle DEF. That is these two triangles ABC and DEF are similar to each other to prove that area of triangle ABC upon area of triangle DEF that is the ratio of the areas of two similar triangle is equal to the ratio of the squares of their corresponding sites that is this would be equal to BC upon ES the whole square equal to AB upon DE the whole square equal to AC upon DF the whole square. Now before starting its proof we have to do some construction work in this. We need to draw AG perpendicular to BC and DH perpendicular to ES. So we have drawn AG perpendicular to BC and DH perpendicular to ES. Now we have the proof now the area of triangle ABC is equal to half into base that is BC into height that is AG. Now similarly the area of the triangle DEF is equal to half into base that is ES into height that is DH. So the ratio of the areas of these two triangles that is area of triangle ABC upon area of triangle DEF is equal to half into BC into AG upon half into ES into DH. So this would be equal to BC into AG upon ES into DH that is area of triangle ABC upon area of triangle DEF is equal to BC into AG over ES into DH. Then we have in triangle ABG and triangle DEH angle B is equal to angle E as we have the triangle ABC is similar to triangle DEF which is already given to us. And then we have and we have angle AGB is equal to angle DH since each is of 90 degrees as AG and DH are the altitudes of the respective triangles. So from the key idea we have that triangle ABG is similar to triangle DEH by AA similarity criterion that we have discussed in the key idea. And therefore from here we have AG upon DH is equal to AV upon DE since the triangles are similar. So the ratio of the corresponding sides would be equal also we have that triangle ABC is similar to triangle DEF since it's already given to us. So from here we have AB upon DE is equal to BC upon EF is equal to CA upon FD. Now we would need this result further so let's take it as 1 then we have this let's take it as 2 let this be 3. Now therefore we have area of triangle ABC upon area of triangle DEF would be equal to BC upon EF into AG upon DH that is from 1. This is further equal to BC upon EF into AB upon DE from 2 since in 2 we have that AG upon DH is equal to AB upon DE. Now this would be further equal to AB upon DE into AB upon DE from 3 since in 3 we have that AB upon DE is equal to BC upon EF. So instead of BC upon EF we have taken AB upon DE that is we get area of triangle ABC upon area of triangle DEF is equal to AB upon DE the whole square. Now using this 3 we get that area of triangle ABC upon area of triangle DEF is equal to AB upon DE the whole square equal to BC upon EF the whole square is equal to AC upon DF the whole square. So we have proved this hence proved. Now we have to do the second part of the question which is that in a triangle ABC XY is parallel to BC and it divides triangle ABC into 2 parts of equal area and we need to prove that BX upon AB is equal to root 2 minus 1 upon root 2. Now let's try doing the second part of the question. Let's see what is given to us in the second part we have a triangle ABC and XY is parallel to BC that is here we have this XY is parallel to the side BC of triangle ABC and this XY divides triangle ABC into 2 parts of equal area. So that we can say that area of AXY is equal to area of BC YX. Now to prove that BX upon AB is equal to root 2 minus 1 upon root 2 first we take that in triangle AXY and triangle ABC we have that angle AXY is equal to angle ABC since they are the corresponding angles and XY is parallel to BC. So the corresponding angles are equal. And angle A is equal to angle A since it is the common angle. So therefore we get that triangle AXY is similar to triangle ABC by AA similarity criterion. This implies that area of triangle AXY upon area of triangle ABC is equal to AX upon AB the whole square using the above result that is the ratio of the areas of two similar triangles is equal to the square of the ratio. Next we have area of triangle ABC would be equal to area of triangle AXY plus area of BC YX. Now it is already given to us that area of triangle AXY is equal to area of BC YX. So we get that area of triangle ABC is equal to two times the area of triangle AXY since we are given that area of triangle AXY is equal to area of BC YX. Thus now we have that area of triangle AXY upon two times area of triangle AXY since we have that area of triangle ABC is equal to two times area of triangle AXY. So putting this value for area of triangle ABC in this we get this and this would be equal to AX upon AB the whole square. So we now get that AX upon AB the whole square is equal to one upon two. So this further implies that AX upon AB is equal to one upon root two. So from this figure we have that AX is equal to AB minus BX. So we have AB minus BX upon AB is equal to one upon root two. That is we have that AB upon AB minus BX upon AB is equal to one upon root two which further gives us one minus BX upon AB is equal to one upon root two. That is we now have BX upon AB is equal to one minus one upon root two which further gives us BX upon AB is equal to root two minus one upon root two. So we have got BX upon AB is equal to root two minus one upon root two hence proved. So this completes the session. Hope you have understood the solution for this question.