Let $E=BC\cap AD$. Easy to see that $E$ is midpoint of $BC$, so $YE\|XC$ and by Reim's we get that $BYEZ$ is cyclic, and it's center is the midpoint of $BE$. Now since $\angle BEA=90^{\circ}$ we have that $AE$ is tangent to $(BYEZ)$, hence, $\angle ZEA=\angle ZYE=\angle ZBE=\angle ZCA \implies ZECA$ is cyclic and the conclusion follows.

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Can you please elaborate on how you used Reim's to find that BYEZ is cyclic? Thank you so much!

tempcm wrote: Can you please elaborate on how you used Reim's to find that BYEZ is cyclic? Thank you so much! You can just do $\angle BEY = \angle BCX = \angle BCD = \angle BZD$ and so $BYEZ$ is cyclic

Let \( R = \frac{1}{2} \), \( \angle BDC = \delta \), and \( \angle YDC = \angle ZCA = \alpha \). Also we have in \( \triangle AOC \) : \( AC = \frac{\tan\left(\delta\right)}{2}\). To prove \( \angle AZC = 90^\circ \), we need to show that \( \cos(\alpha) = \frac{ZC}{AC} = \frac{\sin(\alpha)}{\frac{\tan\left(\delta\right)}{2}} \iff \tan(\alpha) = \frac{\tan\left(\delta\right)}{2}\ \iff \frac{XY}{XD} = \frac{BX}{2XD} \), which is obviously true.

Surprisingly little posts? Also yay one of my first solved hard problems using basics of proj geo We rephrase the problem into proving that AZE is a line where E is the diametrically opposite point of C, since the 90 degrees in a cyclic quad motivates us to draw that diametrical point, and we would be done from CZE=90. Since CDE=90 degrees, DE//BX, and since BY=YX it's well known that D(XYBE) is a harmonic pencil (since projecting onto a line parallel to DE makes intersections B',Y',X',$P_\infty$ with the pencil, with B'Y'/Y'X'=BY/YX=1, now basic cross ratio shows it's harmonic so the pencil is harmonic too). Then the harmonic pencil projected onto a circle yields CZBE a harmonic quad, hence from properties of symmedians' converse ZE and AE are both symmedians of EBC, whence AZE is a line, as desired. $\blacksquare$

huashiliao2020 wrote: Surprisingly little posts? Also yay one of my first solved hard problems using basics of proj geo We rephrase the problem into proving that AZE is a line where E is the diametrically opposite point of C, since the 90 degrees in a cyclic quad motivates us to draw that diametrical point, and we would be done from CZE=90. Since CDE=90 degrees, DE//BX, and since BY=YX it's well known that D(XYBE) is a harmonic pencil (since projecting onto a line parallel to DE makes intersections B',Y',X',$P_\infty$ with the pencil, with B'Y'/Y'X'=BY/YX=1, now basic cross ratio shows it's harmonic so the pencil is harmonic too). Then the harmonic pencil projected onto a circle yields CZBE a harmonic quad, hence from properties of symmedians' converse ZE and AE are both symmedians of EBC, whence AZE is a line, as desired. $\blacksquare$ wait xonks czbe isnt a harmonic quad it's not even cyclic you mean CEBD?

CZBE is a harmonic quad, don't refer to the diagram 2 posts above; E lies on this circle since it's diametrically opposite C. X->C, Y->Z, B->B, E->E from projecting through D onto the circle