A confusion in an IMO geometry problem.

Question

So, I am solving EGMO and had a confusion on this problem;

A circle has center on the side AB of the cyclic quadrilateral $ABCD$. The other three sides are tangent to the circle. Prove that $AD + BC = AB$.

My question is ;

Is it necessary that the center of circle is the midpoint of $AB$? Does it matter?

As I was able to solve most of the other problems, and from what I have heard, this is one of the easier problems so I shouldn't be having much trouble with it but I don't know why before starting the problem (at 2 in the night), I had this confusion in my mind so now I can't move past.

I am sorry if this is a dumb question, but please please clear my confusion.

Answer 1

Answer: No, it's not necessary that the center of the circle is, at the same time, the midpoint of $AB$.

For the proof, consider a point $T$ on $AB$, such that $BT=BC$, and let us employ directed angles. We will proceed to show that $AT=AD$. In fact, let $O$ be the center of the circle, and observe that $DOTC$ is a cyclic quadrilateral too: $$\angle ODC=\frac{\angle ADC}2=\frac{180^\circ-\angle CBA}2=\angle BTC$$ But then $$\angle DTA=\angle DCO=\frac{\angle DCB}2=\frac{180^\circ -\angle BAD}2\implies AT=AD$$

Notice: We did not require $O$ to be the midpoint of $AB$. And playing around (say: with geogebra or desmos) will show that such a configuration is possible.

Answer 2

Attempting a sketch with Geogebra, the answer seems to be no, $O$ does not have to be the centre of $AB$.