Re: 2012 HSC MX1 Marathon
Last Q of one of our mx1 papers ;
Two circles intersect at A and B. P is a point on the first circle and Q is a point on the second circle such that PAQ is a straight line. C is a point on the second circle. The tangent to the first circle at P meets QC produced to R.
Prove that PBCR is a cyclic quadrilateral
Construct BA and point D (I forgot to write point Q which is produced from PA and lies on second circle)
let /_DPB = x
/_PAB = x (angle between tangent and chord equal to angle in alternate segment)
/_BAQ = 180 - x (adjacent supplementary angles)
/_BCQ = 180 - x (angles equal on the same arc)
/_RCB = x (adjacent supplementary angles)
Therefore PBCR is cyclic since angle is equal to opposite external angle /_RCB=/_DPB