A IMO problem about geometry which can probably be solved by simple angle chasing.

Question

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

My Attempt: I extended $DA$ and $BC$ to meet at X.Clearly, The circle is the incircle of $\Delta XCD$ and the line $AB$ is antiparallel to base $CD$. Now Cut $AT$ from $ AB$ where $AT=AD$.Enough to show $BT=BC$.I failed to do anything after this.I believe by finding some cyclic quadrilaterals and by angle chasing it can be done. Please help me.

Answer 1

If $\angle OAD=\alpha$, then $\angle FOC=\alpha/2$ and: $$ AO-AE=r(\csc\alpha-\cot\alpha)=r{1-\cos\alpha\over\sin\alpha}= r\tan{\alpha\over2}=FC, $$ where $r=OE=OF$ is the radius of the inscribed circle. In the same way one derives $BO-BF=ED$ and adding these two equalities we get the desired result.

Answer 2

The angle chase ---which relies upon the fact that opposite angles of an inscribed quadrilateral are supplementary--- gives us this figure:

The result then follows from the identity illustrated in this trigonograph:

$$\tan\theta + \cot 2\theta = \csc 2\theta$$