A point $P$ lies on side $AB$ of a convex quadrilateral $ABCD$. Let $\omega$ be the inscribed circumference of triangle $CPD$ and $I$ the centre of $\omega$. It is known that $\omega$ is tangent to the inscribed circumferences of triangles $APD$ and $BPC$ at points $K$ and $L$ respectively. Let $E$ be the point where the lines $AC$ and $BD$ intersect, and $F$ the point where the lines $AK$ and $BL$ intersect. Prove that the points $E, I, F$ are collinear.
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Tags: geometry
Luis González
28.10.2015 04:25
Let $(U)$ and $(V)$ be the incircles of $\triangle PAD$ and $\triangle PBC.$ Since $(I) \equiv \omega$ and $(U)$ are tangent to $PD$ at the same point $K$ $\Longrightarrow$ $DA+DP-AP=DP+DC-PC$ $\Longrightarrow$ $DA+PC=DC+AP$ $\Longrightarrow$ $APDC$ is tangential with an incircle $\omega_1$ and similarly $BPDC$ is tangential with an incircle $\omega_2.$ Since $A$ is the exsimilicenter of $(U) \sim \omega_1$ and $C$ is the exsimilicenter of $(I) \sim \omega_1,$ then by Monge & d'Alembert theorem, it follows that $X \equiv AC \cap IU$ is the exsimilicenter of $(I) \sim (U)$ and similarly $Y \equiv BD \cap IV$ is the exsimilicenter of $(I) \sim (V)$ $\Longrightarrow$ $XY$ is the positive homothety axis of $(I),(U),(V),$ cutting $UV$ at the eximilicenter $Z$ of $(U) \sim (V).$ Moreover, since $K$ and $L$ are the insimilicenters of $(I) \sim (U)$ and $(I) \sim (V),$ then $Z \in KL.$ Now $\triangle AXK$ and $\triangle BYL$ are perspective through $Z$ $\Longrightarrow$ by Desargues theorem, it follows that $I \equiv XK \cap YL,$ $E \equiv XA \cap YB$ and $F \equiv AK \cap BL$ are collinear.