 Hello and welcome to the session. In this session we discussed the following question which says in the given figure triangle ABC is right-angled at C and DE is perpendicular to AB. Proves that triangle ABC is similar to triangle ADE and hence find the lengths of AE and DE. Before moving on to the solution let's recall the AA similarity criterion. According to this we have 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. Let's proceed with the solution now. We are given this figure let's see what all is given to us in this figure. We have a triangle ABC which is right angle at C that is we have angle ACB is equal to 90 degrees and we have DE is perpendicular to AB then BC is equal to 12 centimeters DC is equal to 2 centimeters AD is equal to 3 centimeters. We need to prove that triangle ABC is similar to the triangle ADE and also we need to find the lengths of AE and DE. So let's start with the proof now. First we will prove that triangle ABC is similar to the triangle ADE. Consider the triangles ABC and ADE in this we have angle ACB is equal to angle AED each equal to 90 degrees since we have triangle ABC is right angle at C so angle ACB is 90 degrees and DE is perpendicular to AB so angle AED is also 90 degrees. Then we have angle BAC is equal to angle DAE since it is the common angle for both the triangles therefore we get that the triangle ABC is similar to the triangle ADE by the AA similarity criterion. Now next we will find the lengths of AE and DE since we know that in similar triangles corresponding sides are proportional therefore AB upon AD is equal to BC upon DE is equal to AC upon AE since we know that triangle ABC is similar to the triangle ADE and thus their sides would be proportional. Now substituting the known values we get AB upon now AD is 3 centimeters so AB upon 3 is equal to BC which is 12 upon DE is equal to AC which is 3 plus 2 5 upon AE. Now let's find out what is AB for this we consider the right triangle ABC in this AB square is equal to BC square plus AC square by the Pythagoras theorem so this means we have AB square is equal to 12 square plus 5 square which means AB square is equal to 144 plus 25 which gives us AB square equal to 169 and from here we get AB equal to 13 centimeters thus putting the value of ABS 13 in this we get 13 upon 3 is equal to 12 upon DE is equal to 5 upon AE. Now considering these two we get DE is equal to 12 into 3 upon 13 that is DE is equal to 36 upon 13 so we get DE is equal to 2.77 centimeters approximately. Now let's consider 13 upon 3 is equal to 5 upon AE this gives us AE is equal to 5 into 3 upon 13 that is 15 upon 13 which gives us AE is equal to 1 point 15 centimeters approximately so this is the value for AE this finally we get AE is equal to 1 point 15 centimeters approximately and DE is equal to 2.77 centimeters approximately so these are the final values for AE and DE this completes the session hope you have understood the solution of this question.