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In the adjoining figure, O is the centre...

In the adjoining figure, O is the centre of the circle, Prove that `anglex=angley+anglez`.

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`angle PBQ=anglePCQ` (angles of same segment)
Arc PQ subtends angles at
centre `=anglePOQ`
remaining circle `=anglePBQ` or `angle PCQ`
`therefore angle PBQ=anglePCQ=(1)/(2)anglePOQ=(z)/(2)`
Now, `angle BDC=angle PDQ=y`
`angle ABD=180^@-anglePBQ=180^@-(z)/(2)`
`angle ACD =180^@-angle PCQ=180^@-(z)/(2)`
In square `ABDC`,
`angle ABD+angle BDC+angle DCA+angle CAB=360^@`
`rArr(180^@-(z)/(2))+y+(180^@-(z)/(2))+x=360^@`
`rArr-z+y+x=0`
`rArrx+y=z`
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