Given
#xcos alpha+ysin alpha-k=0....[1]#
#xcos beta+ysin beta-k=0....[2]#
To show that #x/cos ((alpha+beta)/2)=y/sin ((alpha+beta)/2)=k/cos ((alpha-beta)/2)#
On cross multiplication of [1] an[2] we get
#x/(sinbeta -sin alpha)=y/(cosbeta-cosalpha)=k/(sinbetacosalpha-cosbetasinalpha)#
#=>x/(-2cos((alpha +beta)/2)sin((alpha-beta)/2))=y/(-2sin((alpha+beta)/2)sin((alpha-beta)/2))=k/(-sin(alpha-beta))#
#=>x/(-2cos((alpha +beta)/2)sin((alpha-beta)/2))=y/(-2sin((alpha+beta)/2)sin((alpha-beta)/2))=k/(-2sin((alpha-beta)/2)cos((alpha-beta)/2))#
#=>x/(cos((alpha +beta)/2))=y/(sin((alpha+beta)/2))=k/(cos((alpha-beta)/2))#
