  Question

# ABCD is a quadrilateral such that ∠ABC+∠ADC =180∘ inside the quadrilateral. Statement  1: the circumcircle of ΔABC intersects diagonal BD at D. Statement 2: the circumcircle of ΔABC intersects BD at D′inside the quadrilateral. Statement 3: the circumcircle of ΔABC intersects BD at D′ outside the quadrilateral. Statement 4: the circumcircle of ΔABC does not intersect BD at all. Statement 5: ABCD is called cyclic quadrilateral.Only statement 4 is true  Only statement 1 is true One of the statement 2 or statement 3 can be true Statement 1 and statement 5 are true

Solution

## The correct option is D Statement 1 and statement 5 are true Let us assume the center of the circle is O. suppose the circle intersects BD at D′. We know that the angle subtended by a chord at the center is twice the angle subtended by it at any point on the circle. Now take the line segment AC which is clearly a chord of the circle. As discussed above x = 2∠ABC, y = 2∠ADC. But x and y form a complete angle so x+y = 360∘. So we get ∠ABC + ∠AD′C = 180∘ but given ∠ABC + ∠ADC = 180∘ which can only be satisfied if D and D′ coincide. Thus the circumcircle intersects BD at D itself. As A, B, C, D lie on the circle ABCD is called a cyclic quadrilateral.  Suggest corrections   