Hello snigdha Originally Posted by
snigdha If O is the centre of the circle. Prove that :
< x + < y = < z
I have added the letters A, B and C to your diagram - see the attached.
Then
$\displaystyle \angle QAR = \angle QBR = \tfrac12z$ (angle at centre)
$\displaystyle \Rightarrow \angle PAC = \angle PBC = 180^o-\tfrac12z$ (angles on a straight line)
$\displaystyle \Rightarrow x + y +(180^o-\tfrac12z)+(180^o-\tfrac12z)=360^o$ (angle sum of quadrilateral ACBP)
$\displaystyle \Rightarrow x + y+360^o-z =360^o$
$\displaystyle \Rightarrow x + y = z$
Grandad