The diagonals of cyclic quadrilateral PQRS intersect at U. The
circles's tangent at R meets PS produced at T. If QR = SR, prove
that
PR x ST^2 = UR x RT^2
Haven't been able to sovle this ever since my MYE. Can
someone help?
After 2 hours of work...I finally got a random guess answer
which is most likely partially wrong, but can be used to work
towards the correct answer.
Things you need to know: PR = RT, PSU AND PTR are
similar triangles.
UR / PR = ST / PT
UR / PR = (ST)^2 / (PT)(ST)
UR / PR = (ST)^2 / (PT)(PT - SP)
UR / PR = (ST)^2 / (PT^2 - (PT)(SP))
UR / PR = (ST)^2 / (PT^2 - (1/2(PT) X (PT)))
UR / PR = (ST)^2 / (PT^2 - (1/2)(PT^2))
UR / PR = (ST)^2 / (1/2)(PT)^2
UR / PR = (ST)^2 / (1/2)((RT)^2 + (RP)^2)
UR / PR = (ST)^2 / (1/2)((RT)^2 + (RT)^2)
UR / PR = (ST)^2 / (RT)^2
PR X (ST)^2 = UR X (RT)^2
Discuss any mistakes you find in my workings.