# easy circle geometry.

• Jul 1st 2010, 06:56 PM
earthboy
easy circle geometry.
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)
• Jul 1st 2010, 07:55 PM
simplependulum
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 .
• Jul 1st 2010, 08:42 PM
earthboy
thanks.please posts problems of similar flavour!