Let $ABC$ be an acute triangle with $\angle BAC>\angle BCA$, and let $D$ be a point on side $AC$ such that $|AB|=|BD|$. Furthermore, let $F$ be a point on the circumcircle of triangle $ABC$ such that line $FD$ is perpendicular to side $BC$ and points $F,B$ lie on different sides of line $AC$. Prove that line $FB$ is perpendicular to side $AC$ .
Problem
Source: Baltic Way 2002
Tags: geometry, circumcircle, parallelogram, geometry proposed
jayme
14.11.2010 06:42
Dear Mathlinkers, 1. Note (0) the circumcircle of ABC, (1) the circumcircle of ABD, A' the midpoint of BC, X the second point of intersection of the tangent to (1) at B with (0), (2) the circumcircle of ADF, E the point of intersection of BD and XF. 2. According to a converse of the pivot theorem applied to (0), (1) and (2), E is on (2). 3. According to Reim's theorem applied to (0) and (1), BD // XC. 4. BDA being B-isoceles, BX // ADC ; consequently, BDCX is a parallelogram and A' is the midpoint of DX. 5. The symmetric of (0) wrt BC passing through D, D is the orthocenter of the triangle EFC ; consequently, DC is perpendicular to BF. Sincerely Jean-Louis
Martin N.
14.11.2010 10:55
It is slightly easier to do it the other way round: Take $H$ on $AC$ and $F'\neq B$ on the circumcircle of $\triangle{ABC}$ such that $B$, $F'$ and $H$ lie on a line perpendicular to $AC$. Denote by $G$ the orthocentre of $\triangle{ABC}$ and let $D'$ be the point of intersection of the line parallel to $AG$ through $F'$. As $GH=HF'$, which is well-known, and $D'F'\| GA$, the triangles $\triangle{AGH}$ and $\triangle{D'F'H}$ are congruent, hence $HA=HD'$, implying $AB=BD'$. Consequently, $D'\equiv D$ and $F'\equiv F$, and the result follows.