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Let $ABC$ be a triangle and $DEFG$ be a square, where $D, E$ points are located on $AB$ and $AC$ or their extension line. $F, G$ points are located on $BC$ or the extension of $BC$. The perpendicular distance from $A$ on $BC$ is $2$ units and $BC$ = $6$ units. What is the value of the perimeter of $DEFG$?

Here, I'm little bit confused about the exact location of the points $D, E, F, G$ respectively. I noticed that this 4 points could locate on either on the side of the triangle or on their extension line. So, I thought that the length of the square would be variable according to the various construction of the triangle $ABC$. I couldn't find a way to figure it out. Therefore, I need some help about how to construct that square with keeping its length constant with above mentioned condition.

The error is highly excusable.

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The distance from $A$ to the nearest side of the square is $|x-2|$, where $x$ is a side of the square.

Thus, by the similarity we obtain: $$\frac{|x-2|}{2}=\frac{x}{6},$$ which gives $$x=\frac{3}{2}$$ and the answer $6$ or $$x=3$$ and the answer $12$.

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    $\begingroup$ @Anirban Niloy I don't know to draw in the net. Try to draw the obtuse-angled triangle, $\measuredangle BAC>90^{\circ}.$ The altitude from $A$ is $2$ and $BC=6$. Now, easy to draw our square. $\endgroup$ – Michael Rozenberg Jan 24 at 19:10
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    $\begingroup$ @MichaelRozenberg You assumed that the distance to the nearest side is $2-x$. That means that $DE$ is on the same side of $A$ as $BC$. You also have a solution on the opposite side. The distance is then $x-2$ instead, which yields the solution $x=3$, with perimeter $12$ $\endgroup$ – Andrei Jan 24 at 19:59
  • $\begingroup$ @Andrei I fixed. Thank you! $\endgroup$ – Michael Rozenberg Jan 24 at 20:50

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