 Hello and welcome to the session. In this session we discussed the following question which says, proof that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Using the above result, do the following. In the given figure, DE is parallel to BC and BD is equal to CE, proof that triangle ABC is an isosceles triangle. Let's recall one fact according to which we have triangles on the same base on equal basis and between the same panels are equal in area. This is the key idea to be used in this question. Let's proceed with the solution now. Consider this figure. We are given a triangle ABC in which we have DE is parallel to BC. That is we are given that a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points and we need to prove that the other two sides are divided in the same ratio. That is we have to prove AD upon DB is equal to AE upon EC. First of all, we will join BE and CD. Next we draw DM perpendicular to AC, EN perpendicular to AB. So we have EN perpendicular to AB and DM perpendicular to AC. Next we consider the area of the triangle ADE. This is equal to half into base which is AD into height that is EN. In the same way we have area of the triangle BDE is equal to half into base that is BD into height that is EN. Area of triangle ADE can also be written as half into base. We can take the base now as AE into height that is DM. Then the area of triangle DEC is equal to half into base which is EC into height that is DM. Next we have area of triangle ADE upon area of triangle BDE is equal to half into AD into EN. This hole upon half into BD into EN. Now half half cancels EN EN cancels and this is equal to AD upon BD. That is area of triangle ADE upon area of triangle BDE is equal to AD upon BD. Let this be 1. Next we have area of triangle ADE upon the area of triangle DEC is equal to half into AE into DM. Hold upon half into EC into DM. Half half cancels DM DM cancels and this is equal to AE upon EC that is we have area of triangle ADE upon area of triangle DEC is equal to AE upon EC that this be 2. Now from the figure you can see that triangle BDE triangle DEC are on the same base which is DE between the same parallels that is DE and BC as it is given in the question that DE is parallel to BC. Therefore we see that area of triangle BDE is equal to the area of triangle DEC according to the result stated in the key idea which says the triangles on the same base and between the same parallels are equal in area. Now since area of triangles BDE and DEC are equal therefore from 1 and 2 we get that AD upon BD is equal to AE upon EC or you can say we have AD upon DB is equal to AE upon EC hence proved so we have proved the result. Now the other part of the question says that we are given a triangle ABC where we have DE is parallel to BC and BDE is equal to CE now we have to prove that triangle ABC is an isosceles triangle let's see the proof now triangle ABC DE is parallel to BC so using the above result we have AD upon BDE is equal to AE upon EC we have used this result since we are given that a line is drawn parallel to the side BC and this line intersects the other two sides of the triangle ABC at distinct points D and E so the other two sides are divided in the same ratio thus we get this now since we are given that BD is equal to CE therefore we have AD upon CE is equal to AE upon EC or CE this means we get AD is equal to AE let this be 1 now we are given that BD is equal to EC let this be question 2 then adding equations 1 and 2 we get AD plus BD is equal to AE plus EC so from the figure we have AD plus BD is AB and AE plus EC is AC so we get AB is equal to AC this means that triangle ABC is an isosceles triangle this is what we were supposed to prove so hence proved this completes the session hope you have understood the solution of this question