# Beyond Pythagoras

### Problem

Let $ABC$ be a triangle with circumcenter $O;$ $X$ the intersection of the internal angle bisector of $\angle BAC$ and the perpendicular bisector of $BC.$ $Y$ and $Z$ lie on the external angle bisector of $\angle BAC$ such that $OY\parallel AB$ and $OZ\parallel AC.$ Prove that $[ABZ]+[CAY]=[OBXC]$ where $[W]$ denotes the area of figure $W.$

### Pythagorean theorem

Note that in case $\angle BAC = 90^{\circ}$ the above statement reduces to the Pythagorean theorem. In this case $O$ lies on the hypotenuse $BC$ such that the quadrilateral $BOCX$ degenerates into triangle $BCX;$ and each of the triangles $ABZ,$ $CAY,$ $BCX$ is just a quarter of the square on the corresponding side.

### Solution

By the construction, $AO=BO=CO=OX,$ giving four isosceles triangles $BOX,COX,AOC, AOB.$ For convenience let $\angle BAC=\alpha,$ $\angle AOY=\tau,$ $\angle AOZ=\rho.$ Then, since OZ\parallel AC,\angle CAO=\angle ACO=\rho.$Also, since$YZ$is the external bisector of$\angle BAC,\angle BAZ=\angle CAY=90^{\circ}-\frac{1}{2}\alpha,$implying that$\angle AYO=90^{\circ}-\frac{1}{2}\alpha$also. Now, by the Law of Sines, in$\Delta AOY,\displaystyle\frac{AO}{\sin (90^{\circ}-\frac{1}{2}\alpha )}=\frac{AY}{\sin\tau},$i.e.,$\displaystyle\frac{AO}{\cos\frac{1}{2}\alpha}=\frac{AY}{\sin\tau}.$In$\Delta AOC,AC=2AO\cos\rho.$Finally,$[ACY]=\frac{1}{2}AY\cdot AC\sin (90^{\circ}-\frac{1}{2}\alpha )=\frac{1}{2}AY\cdot AC\cos \frac{1}{2}\alpha.$Combining the three formulas we obtain$[ACY]=AO^{2}\sin\tau\cos\rho.$Similarly,$[ABZ]=AO^{2}\sin\rho\cos\tau.$Now, taking into account that$\tau +\rho =\alpha$and applying the addition formula for sine,$[ACY]+[ABZ]=AO^{2}(\sin\tau\cos\rho +\sin\rho\cos\tau )=AO^{2}\sin\alpha.$To complete the proof, observe that figure$OBXC$combines two isosceles triangles$BOX$and$COX;$the area of each is$\frac{1}{2}AO^{2}\sin\alpha,$because both central angles$BOX$and$COX$equal$\alpha.\$

### Acknowledgment

The problem has been posted by Tran Quang Hung at the CutTheKnotMath facebook page. 