# cyclic quadrilateral

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• Nov 20th 2009, 10:11 PM
thereddevils
cyclic quadrilateral
PQRS is a cyclic quadrilateral where angle PQR and angle QRS are obtuse and have the same value . IF PQ=9 , QR=6 , and RS=4 , prove that if T is the point of intersection of the diagonals , then the circle that passes through R,S and T touches QR at R .
• Nov 21st 2009, 12:45 AM
red_dog
I think it's something wrong with this problem.
If $\widehat{PQR}=\widehat{QRS}=\alpha$ then $\widehat{QPS}=\widehat{RSP}=180-\alpha$

Then $\widehat{QRS}+\widehat{RSP}=180\Rightarrow QR\parallel PS$ and the quadrilater is an isosceles trapezoid. Then $PQ=RS$.

But $PQ=9, \ RS=4$.
• Nov 21st 2009, 10:09 AM
ukorov
If this inscribed quadrilateral has 2 equal obtuse angles, it is most certainly an isosceles trapezoid / trapezium, and so PQ and RS is supposed to have the same length.