 Hello and welcome to the session. In this session we discussed the following question which says in the given figure dEfG is a square and angle BAC is equal to 90 degrees. Show that dE square is equal to VD into EC. Before we move on to the solution, let's recall the AA similarity criterion according to which we have F 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. Move on to the solution now. Consider this figure in which we are given that dEfG is a square then we are also given angle BAC is equal to 90 degrees that is this angle is 90 degrees. We need to show that dE square is equal to BD into EC. Let's proceed with the proof now. First of all let's change the angles in the given figure. Let this be angle 1, this be angle 2, this be angle 3, this be angle 4, this be angle 5 and this be angle 6. Now since dEfG is a square so angle 2 and angle 3 would be of measure 90 degrees since all the angles of a square are of measure 90 degrees. Now let's first consider the triangle BAC in this. We have angle 1 plus angle 4 plus angle 6 is equal to 180 degrees by the angles some property of a triangle according to which we have that the sum of the three angles of a triangle is 180 degrees. Now angle 6 is angle BAC which is of measure 90 degrees so this means angle 1 plus angle 4 is equal to 180 degrees minus 90 degrees that is angle 1 plus angle 4 is equal to 90 degrees. Let this be equation 1. Next we consider the triangle BDG in this triangle we have angle 3 plus angle 4 plus angle 5 is equal to 180 degrees again by the anglesome property of a triangle. Now angle 3 that is this angle is of measure 90 degrees since dEfG is a square so we have angle 4 plus angle 5 is equal to 180 degrees minus 90 degrees that is angle 4 plus angle 5 is equal to 90 degrees. Let this be equation 2. Now from equations 1 and 2 we get angle 1 plus angle 4 is equal to angle 4 plus angle 5 since each is equal to 90 degrees. So this means that angle 1 is equal to angle 5. Let this be equation 3. Now we consider the triangles CEF GDB in these two triangles we have angle 2 is equal to angle 3 each is equal to 90 degrees since dEfG is a square then angle 1 is equal to angle 5 from equation 3. So therefore we get the triangle CEF is similar to the triangle GDB by the AA similarity criterion. So these two triangles are similar now since we know that in similar triangles corresponding sides are proportional therefore the corresponding sides of these two triangles would be proportional since they are similar and so we have CE upon GDB is equal to EF upon DB is equal to CF upon GB. Now let's consider these two ratios that is CE upon GD is equal to EF upon DB. Now since dEfG is a square therefore we can say that CE upon in place of GD we can write ED since all the sides of the square are equal this is equal to time place of EF we can write ED again since all the sides of the square are equal so this is equal to ED upon DB since dEfG is a square so GD is equal to EF is equal to ED. So now cross multiplying we get ED square is equal to DB into CE or you can say dE square is equal to BD into EC and this is what we were supposed to prove. So hence proved this completes the session hope you have understood the solution of this question.