# Rectangular Areas in Circle

### What is this about?

24 January 2015, Created with GeoGebra

### Problem

Two rectangles, $ACDH$ and $BIFC,$ are drawn on the legs of right $\Delta ABC$ such that the points $A,B,F,D$ are concyclic. $DH$ meets circle $(ABFD)$ the second time in point $E,$ $FI$ in point $G.$

Prove that $ABGE$ is a rectangle and $[ABGE]=[ACDH] + [BIFC],$ where $[X]$ denotes the area of shape $X.$

Since $\angle AFG=90^{\circ},$ $AG$ is a diameter of $(ABFD),$ making $\angle ABG$ right. Similarly, $\angle BAE=90^{\circ}.$ Because of the symmetry of chord embedding in a circle, this proves that $ABGE$ is a rectangle.$Let's introduce$a=BC,b=AC,c=AB,x=CD,y=CF,z=AE.$We have to prove that (*)$cz=ay+bx.$There is a short algebraic derivation from the facts, implied in the diagram: 1.$a^{2}+b^{2}=c^{2},$(Pythagorean theorem in$\Delta ABC.)$2.$x^{2}+y^{2}=z^{2},$(Pythagorean theorem in$\Delta AEH.)$3.$ax=by.$(Intersecting chords theorem.) From here$\begin{align} (cz)^{2} &= (a^{2}+b^{2})(x^{2}+y^{2})\\ &= (ax)^{2} + (ay)^{2} + (bx)^{2} + (by)^{2}\\ &= (ax)(by) + (ay)^{2} + (bx)^{2} + (ax)(by)\\ &= (ay)^{2} + 2(ay)(bx)+(bx)^{2}\\ &= (ay+bx)^{2}. \end{align}$Since all quantities involved are positive, this proves (*). ### Acknowledgment The problem above is a reformulation of a cute problem posted by Bùi Quang Tuån at the CutTheKnotMath facebook page. Here's the original problem:$A,B,C,D$are concyclic points on a circle$(O)$and$AC$perpendicular with$BD.$Denote$[X]$the area of$X.$Prove that$\displaystyle\frac{[AED] + [BEC]}{2} = [AOB].\$

A synthetic solution of this problem can be found elsewhere.