Question

ABCD is a quadrilateral such that $\angle$ABC+$\angle$ADC =${180}^{\circ }$. Inside the quadrilateral:

Statement 1: the circumcircle of $\Delta$ABC intersects diagonal BD at D.

Statement 2: the circumcircle of $\Delta$ABC intersects BD at $D^{{}^{\prime }}$inside the quadrilateral.

Statement 3: the circumcircle of $\Delta$ABC intersects BD at $D^{{}^{\prime }}$ outside the quadrilateral.

Statement 4: the circumcircle of $\Delta$ABC does not intersect BD at all.

Statement 5: ABCD is called cyclic quadrilateral.

• A. Statement 1 and statement 5 are true
• B. One of the statement 2 or statement 3 can be true
• C. Only statement 4 is true
• D. Only statement 1 is true

Answer 1

The correct option is A

Statement 1 and statement 5 are true

Let us assume the center of the circle is O. Suppose the circle intersects BD at $D^{{}^{\prime }}$.

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 $\angle$ABC, y=2 $\angle$ADC.

But x and y form a complete angle so x+y= ${360}^{\circ }$.

So we get $\angle$ABC+ $\angle$AD'C=${180}^{\circ }$ but given $\angle$ABC+$\angle$ADC =${180}^{\circ }$ which can only be satisfied if D and $D^{{}^{\prime }}$ coincide.

Thus, the circumcircle intersects BD at D itself.

As A, B, C, D lie on the circle ABCD is called a cyclic quadrilateral.