Question

A, B, C, D are 4 non collinear points in a plane such that $\angle ACB$ =$\angle ADB$ , then how many circle(s) can be drawn passing through all 4 points.

• A. 1
• B. 3
• C. 0
• D. 2

Answer 1

The correct option is A 1

Let us draw the points A, B, C, D such that $\angle ACB$ = $\angle ADB$.

Now draw a circle through A, B, C. let us assume that the circle does not pass through the point D but intersects the extension of line segment CD at $D^{{}^{\prime }}$.

Since angles subtended by an arc in a segment are equal, $\angle ACB$ =$\angle A$ $D^{{}^{\prime }}$B. It is given that $\angle ACB$=$\angle ADB$. Thus for the angles to be equal, D and $D^{{}^{\prime }}$ should coincide. Thus our assumption that the circle does not pass through D is false.

$\therefore$ A circle can be drawn through these 4 points.
So, the given statement is false.