MHB Can AC Be Less Than Twice FG in a Square Geometry Setup?

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In the given geometric configuration involving square ABCD and point E on side AD, the problem requires proving that the length AC is less than twice the length FG. The setup includes point F as the foot of the perpendicular from B to line CE, and point G defined such that BG equals FG, with a specific parallel line condition involving EF. The challenge lies in the relationships between these points and the properties of the square. The discussion emphasizes the geometric reasoning needed to establish the inequality AC < 2·FG. A solution to this problem has not yet been provided, indicating a need for further exploration and analysis.
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Here is this week's POTW:

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Let $ABCD$ be a square, and let $E$ be an internal point on side $AD$. Let $F$ be the foot of the perpendicular from $B$ to $CE$. Suppose $G$ is a point such that $BG = FG$, and the line through $G$ parallel to $BC$ passes through the midpoint of $EF$. Prove that $AC<2\cdot FG$.

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No one answered last week's problem, (Sadface) but you can find the suggested solution as shown below:

View attachment 8747

Note that for given $E$ and $F$, there is only one point $G$ with the required properties.

Since $BG=FG$, $G$ must lie on the perpendicular bisector of $BF$, and by the second condition, $G$ lies on the line through the midpoint of $EF$ parallel to $BC$.

$G$ must thus be the unique intersection of these two lines. We now give an alternative construction for $G$ from which the result will follow.

Since $BA$ and $CD$ are parallel and equal, we may translate $\triangle CDE$ to give a triangle $\triangle BAE'$. Then $E',\,A,\,E$ are collinear, with $E'E=E'A+AE=ED+AE=AD=BC$ and $E'E,\,BC$ are parallel, so we may also translate $\triangle BCF$ to give $\triangle E'EF'$. Then $F,\,E,\,F'$ are collinear. Now, $E'F'=BF$, and $\angle E'F'E=\angle BFC=90^\circ=\angle BFE$. Hence, $BE'F'F$ is a rectangle.

Let $G'$ be its center. Certainly $BG'=FG'$. Let the line through $G'$ parallel to $BC$ hit $BE',\,FF'$ at $M,\,N$ respectively, then symmetry gives $E'M=FN$. However, by translation, $EN=E'M$, so $EN=FN$, and the parallel to $BC$ through $G'$ bisects $EF$. Thus, by the uniqueness of $G$ already proven, $G=G'$, the center of $BE'F'F$.Now, by AM-GM (strict because $BF<BC=CD<CE=BE'$,

$\begin{align*}2\cdot FG&=FE'\\&=\sqrt{BF^2+BE^2}\\&>\sqrt{2\cdot BF\cdot BE'}\\&=\sqrt{2[BFF'E']}\\&=\sqrt{2([BAE']+[E'EF']+[ABFE])}\,\,\,\text{here brackets denote areas}\\&=\sqrt{2([CDE]+[BCF]+[ABFE])}\\&=\sqrt{2[ABCD]}\\&=\sqrt{2}\cdot AB\\&=AC\,\,\,\text{(Q.E.D.)}\end{align*}$
 

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