Tangent circles

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Tangent circles

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Source: Stars of Mathematics 2015 Junior Level #3

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#1 h Jan 2, 2016, 11:29 AM • 1 Y

Y by Adventure10

Let $ABCD$ be cyclic quadrilateral,let $\gamma$ be it's circumscribed circle and let $M$ be the midpoint of arc $AB$ of $\gamma$ ,which does not contain points $C,D$ .The line that passes through $M$ and the intersection point of diagonals $AC,BD$ ,intersects $\gamma$ in $N\neq M$ .
Let $P,Q$ be two points situated on $CD$ ,such that $\angle{AQD}=\angle{DAP}$ and $\angle{BPC}=\angle{CBQ}$ .Prove that circles $\odot(NPQ)$ and $\gamma$ are tangent.

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#2 h Jan 7, 2016, 1:19 AM • 3 Y

Y by enhanced, Adventure10, Mango247

Let $V$ $\equiv$ $AC$ $\cap$ $BD$ and let the tangent of $\gamma$ passing through $N$ cuts the tangent of $\odot (CVD)$ passing through $V$ at $X$ . From $\triangle DAP$ $\sim$ $\triangle DQA,$ $\triangle CBQ$ $\sim$ $\triangle CPB$ we get $\tfrac{{CV}^2}{{DV}^2}$ $=$ $\tfrac{{CB}2}{{DA}^2}$ $=$ $\tfrac{CP \cdot CQ}{DP \cdot DQ}$ , so $VP,$ $VQ$ are isogonal conjugate WRT $\angle CVD$ (Steiner theorem), hence $\odot (PVQ)$ and $\odot (CVD)$ are tangent to each other at $V$ .

Since $VX$ is parallel to the tangent of $\gamma$ passing through $M$ , so $XV$ $=$ $XN$ $\Longrightarrow$ $X$ lies on the radical axis $CD$ of $\odot (CVD)$ and $\gamma$ , hence from ${XN}^2$ $=$ ${XV}^2$ $=$ $XP$ $\cdot$ $XQ$ we conclude that $\odot (PNQ)$ is tangent to $\gamma$ at $N$ .

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