In the given figure, a square is inscribed in a circle with centre O. Find: ∠OCB Is BD a diameter of the circle?
Open in App
Solution
In the given figure we can extend the straight line OB to BD and CO to CA Then we get the diagonal of the square which intersect each other at 90o by the property of Square. From the above statement we can see that ∠COD=90o, The sum of the angle ∠BOC and ∠COD is 180o as BD is a straight line. Hence ∠BOC+∠OCD=∠BOD=180o ∠BOC+90o=180o ∠BOC=180o−90o ∠BOC=90o Thus, triangle OCB is an isosceles triangle with sides OB and OC of equal length as they are the radii of the same circle. in △OCB,∠OBC=∠OCB as they are opposite angles to the two equal sides of an isosceles triangle. Sum of all the angles of a triangle is 180o so, ∠OBC+∠OCB+∠BOC=180o ∠OBC+∠OBC+90o−180o as, ∠OBC=∠OCB 2∠OBC=180o−90o 2∠OBC=90o ∠OBC=45o as ∠OBC=∠OCB So, ∠OBC=∠OCB=45o Yes BD is the diameter of the circle.