, 2017

Since $ABCD$ is a cyclic quadrilateral, $\angle DCB +\angle DAB= 180\degree$ or $(2u)\degree+88\degree=180\degree$ or $2u=92$ and so $u=46$. Since $STQR$ is a cyclic quadrilateral, $\angle SRQ +\angle STQ= 180\degree$ or $x\degree+58\degree=180\degree$ and so $x=122$. Since $PQRS$ is a cyclic quadrilateral, $\angle SPQ +\angle SRQ= 180\degree$ or $y\degree+x\degree=180\degree$ or $y+122=180$ and so $y=58$. In $\triangle JKL$, $KJ=KL$ and so $\angle KLJ=\angle KJL=35\degree$ ($\triangle JKL$ is isosceles). In $\triangle JKL$, $\angle JKL=180\degree-2(35\degree)=110\degree$. Since $JKLM$ is a cyclic quadrilateral, $\angle JML +\angle JKL= 180\degree$ or $\angle JML+110\degree=180\degree$, and so $\angle JML=70\degree$. In $\triangle JLM$, $LJ=LM$ and so $\angle MJL=\angle JML=70\degree$ ($\triangle JLM$ is isosceles). In $\triangle JLM$, $\angle JLM=180\degree-2(70\degree)=40\degree$, and so $w=40$. Solution 1 Since $DEFG$ is a cyclic quadrilateral, $\angle DGF +\angle DEF= 180\degree$ or $\angle DGF+z\degree=180\degree$ and so $\angle DGF=180\degree-z\degree$. Since $FGH$ is a straight angle, then $\angle DGF+\angle DGH=180\degree$ or $(180\degree-z\degree)+\angle DGH=180\degree$ or $\angle {DGH}=180\degree-180\degree+z\degree=z\degree$. Solution 2 Since $DEFG$ is a cyclic quadrilateral, $\angle DGF +\angle DEF= 180\degree$. Since $FGH$ is a straight angle, then $\angle DGF+\angle DGH=180\degree$. Therefore, $\angle DGF+\angle DGH=\angle DGF+\angle DEF$, and so $\angle DGH=\angle DEF=z\degree$.