If $AD \parallel BC,$ then $AD, BC \perp AB$ and ABCD is a rectangle with $AD = BC = \frac{AB}{2}.$ Assume that AC, BC are not parallel and they meet at a point P. Since ABCD is cyclic, $\angle PCD = \angle PAB,\ \angle PDC = \angle PBA$ and the triangles $\triangle ABP \sim \triangle CDP$ are similar. Let $E \in AB$ be the center of the semicircle tangent to AD, BC, CD. (E) is the excircle of the triangle $\triangle CDP$ against the vertex P with exradius $r_p.$ Let (I) be the incircle of the triangle $\triangle ABC$ with inradius r. Denote $p = CD,\ c = DP,\ d = PC$ the sides and $s = \frac{c+ d + p}{2}$ the semiperimeter of the triangle $\triangle CDP.$ Let the bisector $PI \equiv PE$ of the angle $\angle P$ meet the opposite side CD at a point K. The similarity coefficientb of the triangles $\triangle ABC \sim \triangle CDP$ is $\eta = \frac{PE}{PK}.$ The incenter I cuts the angle bisector PK in the ratio $\frac{PI}{IK} = \frac{c + d}{p},$ hence $\frac{PI}{PK} = \frac{c + d}{c + d + p}.$ Also, $\frac{PE}{PI} = \frac{r_p}{r} = \frac{s}{s - p} = \frac{c + d + p}{c + d - p}.$ Thus the similarity coefficient is equal to $\eta = \frac{PE}{PK} = \frac{PE}{PI} \cdot \frac{PI}{PK} = \frac{c + d + p}{c + d - p} \cdot \frac{c + d}{c + d + p} = \frac{c + d}{c + d - p}$ Consequently, $AB = \eta CD = \frac{(c + d)p}{c + d - p}$ $AD = PA - PD = \eta PC - PD = \eta d - c$ $BC= PB - PC = \eta PD - PC = \eta c - d$ $AD + BC = (\eta - 1)(c + d) = \frac{p(c + d)}{c + d - p} = AB$

a) Call $ O$ the center of the circle $ \Gamma$ tangent to $ BC$, $ CD$, $ DA$, and $ P$, $ Q$, $ R$ the respective points where $ \Gamma$ touches $ BC$, $ CD$, $ DA$. Call $ 2\alpha=\angle DAB$, $ 2\beta=\angle CBA$. Triangles $ OPC$ and $ OQC$ are obviously equal and rectangle at $ P$ and $ Q$ respectively, or $ \angle POC=\angle QOC=\frac{\BAD}{2}=\alpha$, Similarly, $ \angle QOD=\angle ROD=\beta$. Calling $ r$ the radius of $ \Gamma$, $ AR=\frac{r}{\tan(2\alpha)}$ and $ CP=r\tan(\alpha)$. Therefore, since $ \frac{1}{\tan(2\alpha)}+\tan(\alpha)=\frac{\cos(2\alpha)+2\sin^{2}\alpha}{\sin(2\alpha)}=\frac{1}{\sin(2\alpha)}=\frac{OA}{r}$, then $ AR+CP=OA$, and similarly $ BP+DR=OB$, or $ AD+BC=AB$, qed. b) If $ BC$ and $ AD$ are parallel, so are $ OP$ and $ OR$, or $ BC$ and $ AD$ are perpendicular to $ AB$. The area of $ ABCD$ is then $ \frac{AB(BC+DA)}{2}=\frac{x^{2}}{2}$ (incidentally, $ ABCD$ is then a rectangle or it could not be cyclic, and $ AB=CD=2BC=2DA$). If $ BC$ and $ AD$ are not parallel, let $ E$ be the point where they meet. Since $ ABCD$ is cyclic, $ ABE$ and $ DCE$ are similar, with proportionality constant $ \frac{AB}{CD}=\frac{x}{y}$, or the area of $ ABCD$ is equal to $ 1-\frac{y^{2}}{x^{2}}$ times the area of $ ABE$. Let us now show that the perimeter of $ ABE$ is determined by $ x$ and $ y$: it is obvious that $ DE+CE=\frac{y}{x}(AE+BE)=AE+BE-(AD+BC)=AE+BE-x$, or $ AE+BE+AB=\frac{x^{2}}{x-y}+x$. Since $ AB$ is given, the area will reach a maximum when $ ABE$ is isosceles in $ E$, ie, when $ ABCD$ is an isosceles trapezoid with $ AB$ parallel to $ CD$, and $ BC=AD=\frac{x}{2}$. Then, the altitude from $ C$ or $ D$ onto $ AB$ is given by $ \sqrt{AD^{2}-\left(\frac{x-y}{2}\right)^{2}}=\frac{1}{2}\sqrt{2xy-y^{2}}$. Multiply this by $ \frac{x+y}{2}$ and the maximum area of $ ABCD$ is found.

Part a) is IMO 1985 I know one very nice proof of this part(without using trigonometry) If someone want i could post it.

Let $ O$ be the center of the circle centered on $ AB$ and tangent to $ AD,DC,CB.$ Rays $ OC,OB$ bisect $ \angle BCD = \theta$ and $ \angle CDA = \varphi.$ Pick the point $ M$ on $ AB$ such that $ AD = AM.$ Then the triangle $ \triangle ADM$ is isosceles, where $ \angle DMA = \frac {\pi - \angle DAB}{2} = \frac {\pi}{2} - \frac {\pi - \theta}{2} = \frac {\theta}{2}$ $ \angle DCO = \frac {\theta}{2}$ $ \Longrightarrow$ $ DCOM$ is cyclic $\Longrightarrow$ $ \angle CMB = \angle ODC = \frac {\varphi}{2}$ On the other hand, in the $ \triangle CMB,$ we have $ \angle MCB = \pi - \frac {\varphi}{2} - (\pi - \varphi) = \frac {\varphi}{2} \Longrightarrow \angle CMB = \angle MCB = \frac {\varphi}{2}$ $ \triangle CMB$ is then isosceles with apex $ B$ $ \Longrightarrow$ $ AB = AM + BM = AD + BC.$

Erken wrote: Part a) is IMO 1985 I know one very nice proof of this part(without using trigonometry) If someone want i could post it. Yes it's the circle centered at A with the length AD then we show that the second triangle is isosceles by some angle chasing and we are done ...

Fouad-Almouine wrote: Erken wrote: Part a) is IMO 1985 I know one very nice proof of this part(without using trigonometry) If someone want i could post it. Yes it's the circle centered at A with the length AD then we show that the second triangle is isosceles by some angle chasing and we are done ... Yes, I like this solution... For who wants to see this solution here (first solution) : https://artofproblemsolving.com/community/c6h60782p366584

Could someone explain why in part b of daniel73 solution it is needed to show that the perimeter of $ ABE$ is determined by $ x$ and $ y$? Thanks