 Hello and welcome to the session. The given question says, in figure 1, D is the point on site BC of triangle APC such that angle ADC is equal to angle BAC. Prove that CA divided by CD is equal to CB divided by CA. So, here we are given a triangle ABC in which angle ADC is equal to angle BAC where D is a point on the site BC and we have to prove that CA divided by CD is equal to CB divided by CA. Let us now start with the proof. Now in triangle A, CD and triangle BAC we are given that angle ADC is equal to angle BAC, angle C is equal to angle C which is in both the triangles and is common therefore angle C is equal to angle C by A similarity. We can say that triangle ADC is similar to triangle BAC and this implies that CA divided by CD is equal to CB divided by CA. Since if in two triangles the corresponding angles are equal then their corresponding sites are proportional. Thus we have proved what we are required to prove. This completes the session. Bye and take care.