Prove that a point is NOT the centre of a circle

Question

Prove that a point is NOT the centre of a circle

I understand that angle DAX = 54 by the alternate segment theorem and so angle ADX = 32

Moreover, I understand that IF X IS the centre, then angle CDX must equal 36... However, I do not know how to show this is not the case.

Answer 1

You may proceed by the method of contradiction (reductio ad absurdum).

First assume that $X$ is the centre.

This implies that $XB = XC$ are both radii to the circle. This makes $\triangle XBC$ isosceles, so $\angle XBC = \frac 12(180 - 94)^{\circ} = 43^{\circ}$ (angle sum of triangle).

Furthermore, since $DXB$ is a straight line segment that forms the diameter of the circle, $\angle DCB = 90^{\circ}$ (angle in a semicircle).

This implies $\angle BDC = (180 - (43+90))^{\circ} = 47^{\circ}$ (angle sum of triangle).

This gives $\angle BDF = \angle BDC + \angle CDF$ (adjacent angles) $= (47+54)^{\circ} = 101^{\circ}$.

But $\angle BDF = 90^{\circ}$ (tangent $DF$ perpendicular to radius $XD$)

Hence a contradiction has been arrived at, and the original assumption is false. Ergo, $X$ is not the centre of the circle ($QED$).

Answer 2

If $X$ is the centre, then $\angle CXB=2\angle CDB$ and $\angle CDB+\angle FDC=\angle FDX=90^\circ$.