# Inspired by Morley's Theorem

Geometry: Auxiliary Lines: Geometric solution by congruence , cyclic quadrilaterals , similarity , homothety (without trigonometry) .

As shown in the figure: Find $X,Y,Z$

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Some considerations: $CP\perp AB$, $AN\perp BC$. If we call $R=AN\cap CP$, $R$ is the orthocenter of $ABC$, so $BR\perp AC$. Since $\sin 10° \sin 40° \sin 50° = \sin 20° \sin 30° \sin 30°$, by the Trig Ceva Theorem we have that $BM\perp AC$, so $B,R,M$ are collinear on the height relative to the $AC$ side. – Jack D'Aurizio Nov 9 '12 at 10:56
Moreover, $RBN$ and $RBP$ are isosceles triangles, so $R$ is the circumcenter of $BNP$ and $RN=RP$. – Jack D'Aurizio Nov 9 '12 at 11:42
By appling the Sine Theorem to triangles $BCM$ and $BCP$, since $\sin^2 50°-\sin^2 30°=(2\sin 20°\sin 80°)^2-(2\sin 10°\sin 80°)^2$, we have that $CM^2-BM^2=CP^2-BP^2$, so $PM\perp BC$. In particular, $AN$ and $PM$ are parallel lines, so $Z=\widehat{RNP}=40°$. – Jack D'Aurizio Nov 9 '12 at 14:01
Now, since $\widehat{NRM}=\widehat{NPM}=Z=40°$, $NRPM$ is a cyclic quadrilateral, so $X=80°$ and $Y=60°$. – Jack D'Aurizio Nov 9 '12 at 14:05
@Sawarnik: done. :) – Jack D'Aurizio Feb 22 '14 at 11:10

Consider that $CP\perp AB, AN\perp BC$. If we call $R=AN\cap CP$, $R$ is the orthocenter of $ABC$, so $BR\perp AC$. Since $\sin10°\sin40°\sin50°=\sin20°\sin30°\sin30°$, by the Trig Ceva Theorem we have that $BM\perp AC$, so $B,R,M$ are collinear on the height relative to the $AC$ side.
Moreover, $RBN$ and $RBP$ are isosceles triangles, so $R$ is the circumcenter of $BNP$ and $RN=RP$.
By appling the Sine Theorem to the triangles $BCM$ and $BCP$, since $\sin^2 50°−\sin^2 30°=(2\sin 20°\sin80°)^2−(2\sin10°\sin80°)^2$, we have that $CM^2−BM^2=CP^2−BP^2$, so $PM\perp BC$. In particular, $AN$ and $PM$ are parallel lines, so $\widehat{Z}=\widehat{RNP}=40°$.
Now, since $\widehat{NRM}=\widehat{NPM}=\widehat{Z}=40°$, $NRPM$ is a cyclic quadrilateral, so $X=80°$ and $Y=60°$.