Geometry

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Rid1
Posts:2
Joined:Sat Aug 26, 2017 3:41 am
Geometry

Unread post by Rid1 » Sat Aug 26, 2017 8:59 pm

Four points P,Q,R and S are taken from the circumference of the circle in anticlockwise direction such that QR>PQ and SP>RS. The bisector oof <PQR and <RSP intersect the circle at point M and N respectively. In the cyclic hexagon drawn by the points P, Q, R, S, M, N four sides are equal to each other. If PM=5, then SQ= ?

samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm

Re: Geometry

Unread post by samiul_samin » Sun Feb 17, 2019 11:49 am

Rid1 wrote:
Sat Aug 26, 2017 8:59 pm
Four points $P,Q,R ,S$ are taken from the circumference of the circle in anticlockwise direction such that $QR>PQ$ and $SP>RS$. The bisector of $\angle{PQR}$ and $\angle{RSP}$ intersect the circle at point $M$ and $N$ respectively. In the cyclic hexagon drawn by the points $P, Q, R, S, M, N$ four sides are equal to each other. If $PM=5$, then $SQ=$ ?

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