 Hi and welcome to the session. Let's work out the following question. The question says, O D is perpendicular to a chord A B of a circle whose center is O. If B C is a diameter, prove that C A is equal to twice of O D. So let this be the circle given to us with center O. Now O D is perpendicular to the chord A B of this circle. B C is given to be the diameter. So let us start with a solution to this question. We have to prove that C A is equal to twice of O D. Now first of all, we join A to C and now we can start with the proof. Since O D is perpendicular to A B level, A D is equal to B D. Because we see that perpendicular drawn on any chord from the center of the circle bisects that chord, it means that D is the midpoint of A B and O C is equal to O B because they are the radiate. Therefore O is the midpoint of B C. Now in triangle B A C, D and O, they are midpoints of A B and B C respectively. Therefore O D is equal to half of A C because line joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it. So we can say that A C is equal to twice of O D or we can say that C A is equal to twice of O D. So this is what we were supposed to prove in this question. I hope that you understood the solution and enjoyed the session. Have a good day.