Circle Geometry Question (Past HSC) (1 Viewer)

laters · Jun 17, 2015

Hi, could someone please try the last part? I've basically given too much time to this question so it may just be that I missed something completely. The only thing I have done that might be of use after successfully completing parts i to iv was to prove that ZYOC was also cyclic. It's from the 1994 HSC.

InteGrand · Jun 17, 2015

laters said:
Hi, could someone please try the last part? I've basically given too much time to this question so it may just be that I missed something completely. The only thing I have done that might be of use after successfully completing parts i to iv was to prove that ZYOC was also cyclic. It's from the 1994 HSC.

View attachment 32126

Once you've proven that ZYOC is cyclic,

$$\angle ZYO = 180^\circ - \angle BCO$ (opposite angles of cyclic quadrilateral $ZYOC$)$

$$=180^\circ - \angle BCP$ (as $\angle BCP = \angle BCO$ as $CPO$ is a straight line)$

$$=180^\circ - \angle XYM$ (as $\angle BCP = \angle XYM$ by the previous part of the question)$

$\Rightarrow \angle ZYO + \angle XYM =180^\circ$

$$\Rightarrow XYZ$ is a straight line (as the adjacent angles $ZYO$ and $XYO$ are supplementary (as $\angle XYO = \angle XYM$, as $YOM$ is a straight line)). \blacksquare$

Circle Geometry Question (Past HSC) (1 Viewer)

laters

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InteGrand

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