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. True
• B. False
• C. 2
• D. 3

Answer 1

The correct option is B

False

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

Now draw a circle trough 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.