The circumference inscribed on the triangle $ABC$ is tangent to the sides $BC$, $CA$ and $AB$ on the points $D$, $E$ and $F$, respectively. $AD$ intersect the circumference on the point $Q$. Show that the line $EQ$ meet the segment $AF$ at its midpoint if and only if $AC=BC$.
Problem
Source: Spanish Communities
Tags: geometry unsolved, geometry
Arne
17.04.2006 01:53
Let $M$ be the intersection of $EQ$ and $AB$. We have \begin{eqnarray*} & & AM = MF \\ & \iff & AM^2 = MF^2 \\ & \iff & AM^2 = MQ \cdot ME \\ & \iff & \Delta AMQ \sim \Delta EMA \\ & \iff & \angle AEM = \angle QAM \\ & \iff & \angle EDA = \angle DAB \\ & \iff & DE \parallel AB \\ & \iff & \angle EDC = \angle ABC \\ & \iff & 90^{\circ} - \frac{1}{2} \angle ACB = \angle ABC \\ & \iff & \angle ACB + 2 \angle ABC = 180^{\circ} \\ & \iff & \angle ABC = \angle BAC \\ & \iff & AC = BC. \blacksquare \end{eqnarray*}
yoodotan0424
21.10.2013 11:07
By harmonic pencil its obvious
PlatinumFalcon
23.11.2014 14:28
Clearly $QEDF$ is harmonic, so take pencil $E(QEDF)$ and intersect it with $AB$. We must have that $(AFQ\infty)=(AFQ ED\cap AB)$, so $ED$ is parallel to $AB$. As $EDC$ is isosceles, the result follows. $\Box$
baenanabread
06.08.2021 21:46
$$AM \cong FM$$$$\Leftrightarrow AM^2 = MQ \cdot ME$$$$\Leftrightarrow \triangle MQA \sim MAE$$$$\Leftrightarrow \angle MAQ \cong \angle AEM$$$$\Leftrightarrow \angle FAD \cong \angle AEQ$$$$\Leftrightarrow \angle BFD - \angle QDF = \angle AEF - \angle QEF$$$$\Leftrightarrow \alpha + \gamma - \angle QDF = \beta + \gamma - \angle QDF$$$$\Leftrightarrow 2\alpha = 2\beta$$$$\Leftrightarrow \angle BAC \cong \angle CBA$$$$\Leftrightarrow BC \cong AC$$$$\square$$
Albert123
14.12.2021 18:31
$\rightarrow$ Let $M$ be the midpoint of AF $(A,L;Q,D) \overset{E}{=}(A,F;M,DE \cap AF) \implies DE \parallel AF \implies AC=BC$. $\leftarrow$ $AC=BC \implies ED \parallel AF$ Note that: $(A,L;Q,D)=-1 \overset{E}{=}(A,F;M,ED \cap AF=\infty_{AF}) \implies AM=MF$. and we finished .