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Geometry Difficulty 3.0 AMC 10/12 Prove it Canada

A cyclic quadrilateral is a quadrilateral whose four vertices lie on some circle. In a cyclic quadrilateral, opposite angles add to $180^{\circ}$ . In the diagram, $ABCD$ is a cyclic quadrilateral. Therefore, $\angle ABC+\angle ADC=180^{\circ}=\angle BAD+\angle BCD$ .

In Figure A below, $ABCD$ is a cyclic quadrilateral.

If $\angle BAD=88^{\circ}$ , what is the value of $u$ ?
In Figure B, $PQRS$ and $STQR$ are cyclic quadrilaterals.

If $\angle STQ=58^{\circ}$ , what is the value of $x$ and what is the value of $y$ ?
In Figure C, $JKLM$ is a cyclic quadrilateral with $JK=KL$ and $JL=LM$ .

If $\angle KJL=35^{\circ}$ , what is the value of $w$ ?
In Figure D, $DEFG$ is a cyclic quadrilateral. $FG$ is extended to $H$ , as shown.

If $\angle DEF=z^{\circ}$ , determine the measure of $\angle DGH$ in terms of $z$ .

Solution

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 $\angle DGH=\angle DEF=z\degree$ .

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