Tuesday, March 16, 2010

Fox 264

Die no soon, before all your story are told.
--- Yoda Fox

1 comment:

  1. Let PQRS be the quadrilateral with P at the top and going counterclockwise. Let PR and QS intersect in M. Now since triangles PQS and QRS are similar, we've A/D = (PQ/RS)^2 = (QM/MS)^2
    while QS=2PQsin(θ/2) and RS=2QSsin(θ/2)=4PQ[sin(θ/2)]^2. Now B/A = SM/MQ = RS/PQ = 4PQ[sin(θ/2)]^2/PQ = 4[sin(θ/2)]^2 -----(2)
    RM/MP=RS/PQ=C/A=4[sin(θ/2)]^2=B/A. Hence, B=C----(1)
    D/A =(RS/PQ)^2=16[sin(α/2)]^4---(3)
    Now A/B = QM/MS=PQ/RS=PM/MR=B/D or B^2=AD--(4)