 Hello and welcome to the session. My name is Mansi and I'm going to help you with the following question. The question says in figure 8.31 BD is perpendicular to AC and CE is perpendicular to AB. Prove that first triangle AEC is similar to triangle ADB and second CA divided by AB is equal to CE divided by DB. So in this question we need to prove two things this and this. Let us start with the solution to this question. First of all we have to show that triangle AEC is similar to triangle ADB. So in triangle AEC and triangle ADB. First of all we see that angle A is common in both the triangles. So first thing is that angle A is equal to angle A because that is a common angle. Second thing that we see is angle BDA is equal to angle CEA because both of them are equal to 90 degrees because these are perpendicular as we see it's given the question. And third we see that angle ACE is equal to angle ABD because this is the third angle. If two angles of one triangle are equal to corresponding two angles of other triangle then third angle is also equal to third angle of the other triangle. So from these three we have triangle AEC is similar to triangle ADB by angle angle angle similarity criterion. Now we need to prove the second part that means we need to prove that CA by AB is equal to CE by DB. Now in these two triangles we see that CA by AB will be equal to CE by DB. This we get by we see that these two triangles are similar so their corresponding sides will be in ratio therefore we have this. Hence proved. So I hope that you understood the question and enjoyed the session. Have a good day.