In right-angle triangle $ABC$, $\angle C=90$°, Point $D$ is the midpoint of side $AB$. Points $M$ and $C$ lie on the same side of $AB$ such that $MB\bot AB$, line $MD$ intersects side $AC$ at $N$, line $MC$ intersects side $AB$ at $E$. Show that $\angle DBN=\angle BCE$.
Problem
Source: China south east mathematical Olympiad 2007 problem 6
Tags: geometry, circumcircle, cyclic quadrilateral, geometry unsolved
Luis González
12.07.2013 07:35
Let $F \equiv AC \cap MB.$ Circle $(O)$ with diameter $\overline{BF}$ is obviously orthogonal to the circumcircle of $\triangle ABC$ $\Longrightarrow$ $DB$ and $DC$ are tangents of $(O).$ Let $G$ be the second intersection of $CM$ with $(O)$ and $H \equiv FG \cap BC.$ From the complete cyclic quadrilateral $CFGB,$ it follows that $M,$ $FC \cap BG$ and $D$ are on the polar of $H$ WRT $(O)$ $\Longrightarrow$ $N \equiv FC \cap BG$ $\Longrightarrow$ $\angle DBN \equiv \angle EBG=\angle BCE.$
sunken rock
12.07.2013 14:34
Let the parallel through $N$ to $AB$ intersect $CE, BM, BC$ at $P, R, Q$ respectively and call $S\in BN\cap CE$. Since $D$ is midpoint of $AB$ and $NP\parallel AB$ it follows that the pencil $N(A, D, B, P)$ is harmonic, and so are $B(C, M, S, P)$ and $B(Q, R, N, P)\ (\ 1\ )$. Obviously, $F$ is the orthocenter of $\Delta BNQ$, and from $(1)$ we infer $Q-F-S$ collinear, hence $FS\bot BN$ and $CBSF$ is cyclic, done. Best regards, sunken rock
