easy circle geometry.

Feb 2010
163
15
in the 4th dimension
ABCD is a cyclic quadrilateral.AB and DC when produced meet at P.again AD and BC when produced meet at R.The circumcircles of triangles BCP and CDR intersect at T. prove that P,T,R are collinear.
i have a soloution,but i want to use the Menelaus theorem or some trignometry to prove it. Is it possible to do so???(Wondering)
please HELP!
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
I cannot think of the method using Menelaus Theorem , seemingly it is very difficult for us to apply it ...


\(\displaystyle \angle RDT = \angle RCT = \angle BPT \)

\(\displaystyle \angle CDT = \angle CRT = \angle BRT\)

Therefore \(\displaystyle \angle ADC = \angle RDT + \angle CDT = \angle BPT + \angle BRT \) also \(\displaystyle \angle ADC = \angle ABR \) so we have


\(\displaystyle \angle BPT + \angle BRT = \angle ABR \)

Therefore \(\displaystyle P,T,R \) lie on a straight line .
 
Feb 2010
163
15
in the 4th dimension
thanks.please posts problems of similar flavour!
 
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