USA IMO

First Solution. Let $\alpha$ be the angle formed by diagonals $AC$ and $BD$, where $0^{\circ }<\alpha \le 90^{\circ }$. We calculate the area of quadrilateral $ABCD$ in two ways. We obtain $$[ABCD]=\frac{1}{2}(AC\cdot BD\sin \alpha )$$ and $$\begin{array}{rl}[ABCD]& =[ABC]+[CDA]\\ & =\frac{1}{2}(AB\cdot BC\sin 135^{\circ })+\frac{1}{2}(CD\cdot DA\sin 135^{\circ })\\ & =\frac{1}{2\sqrt{2}}(AB\cdot BC+CD\cdot DA).\end{array}$$ Combining the above two equations yields $$A{C}^2\cdot B{D}^2{\sin }^2\alpha =\frac{1}{2}(AB\cdot BC+CD\cdot DA{)}^2.$$ For real numbers $a$ and $b$, $(a+b{)}^2\ge 4ab$, hence $$A{C}^2\cdot B{D}^2{\sin }^2\alpha \ge 2AB\cdot BC\cdot CD\cdot DA.$$ It follows that ${\sin }^2\alpha \ge 1$. Therefore, $\sin \alpha =1$ and $\alpha =90^{\circ }$, as desired.

Second Solution. Consider an inversion of center $B$ and radius $1$. Let $A^{\prime },C^{\prime },D^{\prime }$ denote the images of $A,C,D$ under the inversion, respectively. Then $\angle A^{\prime }{D}^{\prime }B=\angle BAD$ and $\angle B{D}^{\prime }{C}^{\prime }=\angle DCB$. It follows that $$\angle A^{\prime }{D}^{\prime }{C}^{\prime }=\angle BAD+\angle BCD=360^{\{}\circ \}-2\cdot 135^{\{}\circ \}=90^{\{}\circ \}.$$ Applying the Inversive Distance Formula we obtain $$A^{\prime }{C}^{\prime }=\frac{AC}{AB\cdot BC},C^{\prime }{D}^{\prime }=\frac{CD}{BD\cdot BC},D^{\prime }{A}^{\prime }=\frac{DA}{AB\cdot BD}.$$ Applying the Pythagorean Theorem to right triangle $A^{\prime }{C}^{\prime }{D}^{\prime }$ yields $${\left(\frac{AC}{AB\cdot BC}\right)}^2={\left(\frac{CD}{BD\cdot BC}\right)}^2+{\left(\frac{DA}{AB\cdot BD}\right)}^2,$$ or $$(AC\cdot BD{)}^2=(AB\cdot CD{)}^2+(BC\cdot DA{)}^2.$$ Rewriting this as $$(AB\cdot CD-BC\cdot DA{)}^2=0,$$ we have $AB\cdot CD=BC\cdot DA$; that is, $AB/BC=AD/CD$. Because $\angle ABC=\angle CDA$, it follows that triangles $ABC$ and $ADC$ are similar. These two triangle share the corresponding side $AC$, so they are congruent. This implies that $ABCD$ is a kite with segment $AC$ as its symmetry axis. Therefore, $AC\perp BD$.