Feels too straightforward. Am I missing something here? YaWNeeT wrote: Given a circle and four points $B,C,X,Y$ on it. Assume $A$ is the midpoint of $BC$, and $Z$ is the midpoint of $XY$. Let $L_1,L_2$ be lines perpendicular to $BC$ and pass through $B,C$ respectively. Let the line pass through $X$ and perpendicular to $AX$ intersects $L_1,L_2$ at $X_1,X_2$ respectively. Similarly, let the line pass through $Y$ and perpendicular to $AY$ intersects $L_1,L_2$ at $Y_1,Y_2$ respectively. Assume $X_1Y_2$ intersects $X_2Y_1$ at $P$. Prove that $\angle AZP=90^o.$ Proposed by William Chao Invert at $A$. We won't use $'$s to denote images. It's too edgy Let $T=\overline{BX} \cap \overline{CY}$ and $S=\overline{BY} \cap \overline{CX}$. Then $\overline{X_1Y_2} \mapsto \odot(AT)$ while $\overline{X_2Y_1} \mapsto \odot(AS)$. Finally, $AXZY$ is harmonic; we need to show $\odot(AZ), \odot(AS), \odot(AT)$ are coaxial, or $Z$ lies on $\overline{ST}$. However this follows by noting that they all lie on the polar of $\overline{XY} \cap \overline{BC}$ in $\odot(BXCY)$.

@above, I can’t see why $X_{1}Y_{2}$ is transformed like this... And I don’t think $S,T,Z$ is collinear since $Z$ is the midpoint of $XY$. Can you explain it explicitly?

It's pretty normal to try out inverting at $A$ since lots of circle passes through that point, but let me try not using any inversion Let $M,N$ be the foot of perpendicular from $A$ to $X_1Y_2$ and $X_2Y_1$ respectively. We could easily verify that $ACXX_2N, ABY_1NY,ABXX_1M,ACYY_2M$ are cyclic Radical axis of $(ACXX_2N), (ABY_1NY),(BCXY)$ give $AN,CX,BY$ concurrent denotes that concur point by $S$. Similarly $AM, BX, CY$ are concurrent at point called $R$ Now, looking at $(APMN),(BCXY),(ACXX_2N)$ gives that radical axis of $(BCXY),(APMN)$ passes through $S$, and similarly we could prove that it passes through $R$ Thus, $SR$ is the radical axis of $(BCXY),(APMN)$, let $SR$ cuts $BC$ at $E$ and $XY$ at $F$ let $BC$ cuts $XY$ at $Q$ Since $(Q,E : B,C) = 1$ , we have $EB \cdot EC = EA \cdot EQ$ but $E$ lies on the radical axis of $(BCXY),(APMN)$ hence, $Q$ lies on $(APMN)$ Now, since $(Q,F : X,Y) = 1$ we have $FQ \cdot FZ = FX \cdot FY$ but $F$ lies on the radical axis of $(BCXY),(APMN)$ hence $Z$ lies on $(APMN)$ Thus, $\angle{AZP} = \angle{AMP} = \angle{ANP} = 90^o$ as desired

Since there are midpoints of chord, draw the center of $(BCXY)$ say $O$. Let's handle the overloading information a bit. 1) $BAXX_1$ and $CAYY_2$ are cyclic. These two circles meet at the feet of altitude of $A$ to $X_1Y_2$, say $U$. 2) Define $V$ as the feet of altitude of $A$ to $X_2Y_1$, then $V$ lies on $(BAYY_1), (CAXX_2)$ too. 3) So $AUVP$ is cyclic with diameter $AP$. I will now show that $Z$ and $O$ both lies on $(AUVP)$ too, which I will only show $AUOZ$ is cyclic. Then similarly $AVOZ$ will be cyclic then $Z$ will lie on the circle of diameter $AP$. I reduce it to the following: Let $ABCD$ be a cyclic quadrilateral with $E, F$ are midpoints of $AB, CD$. Then if $(EBC)\cap (EAD)=K$ and $O$ is the center of the circle $(ABCD)$, then $OKEF$ is cyclic. Apply an inversion about $(O)$ to get the following: Let $ABCD$ be a cyclic quadrilateral with $E, F$ are poles of $AB, CD$. Then if $(EBC)\cap (EAD)=K$, then $K$ lies on $EF$ 4) $EK$ is radical axis of $(EBC), (EAD)$, and $G=BC\cap AD$ lies on it. 5) But we know $G, E, F$ are colinear already, so $K$ lies on $EF$. $\blacksquare$