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Let BC touch in circle at X & AC at Y. Let AB touch excircle at Z.Let BD = p & FX = qBDIX is a square of side p & BFEZ is a square of side p+qNow DZ = 2p+q = YG = q + 2.CF, so CF = CG = pHence Tr. s BDF & GCE are congruent SAS and hence < BFD = < CEG = < CFG since CFEG is concyclic It follows that D,F,G are collinearSumith PeirisMoratuwaSri Lanka
Draw DIP diameter. Ang APD = ACB = 2.CGF => PDG = CGFbut PDG = DF'B => DF'B = CFG (CF=CG equal tangencial seg)
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Let BC touch in circle at X & AC at Y.
ReplyDeleteLet AB touch excircle at Z.
Let BD = p & FX = q
BDIX is a square of side p & BFEZ is a square of side p+q
Now DZ = 2p+q = YG = q + 2.CF, so CF = CG = p
Hence Tr. s BDF & GCE are congruent SAS and hence < BFD = < CEG = < CFG since CFEG is concyclic
It follows that D,F,G are collinear
Sumith Peiris
Moratuwa
Sri Lanka
Draw DIP diameter. Ang APD = ACB = 2.CGF => PDG = CGF
ReplyDeletebut PDG = DF'B => DF'B = CFG (CF=CG equal tangencial seg)