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Let $\Delta ABC$ be a right angled isosceles triangle with $\angle A=90^°$. $D$ is the foot of its altitude from $A$, $ M$ the midpoint of $AD$ and $S$ the midpoint of $AM$.

$B'$ is the reflection of $B$ across the line $CM$, $C'$ is the reflection of $C$ across the line $BM$.

Let E be the intersection of $CB'$ and $BC'$.

Construct a parallel to $BC$ through the point $A$, denote by $P$ its intersection with segment $CE$ by. A parallel line to $BC$ through $S$ meets $AC$ at the point $Q$.

Show that the quadrilateral $ASQP$ is a square.

It is easy to see that triangle $\Delta EBC$ is isosceles and that $ED$ coincides with $AD$. From here I managed to prove this by using some trigonometric formulas and I was wondering if there was a synthetic way of showing this fact.

The only problem that I have is showing that $ASQP$ is a parallelogram.

Ban
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2 Answers2

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Here is a hint toward a synthetic solution.

enter image description here

Note that with the added lines and vertices, you can describe several similar right triangles, and the proportion of the legs of one of these triangles is easily found. You should then be able to prove that $|AS|=|AP|=|SQ|$.

David K
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Since there is no circle in your figure, and we are dealing with a right-angled isosceles triangle, placing the points on the coordinate system seems to be an effortless approach.

WLOG, let's assume $A=(0,2), B=(-2,0)$, and $C=(2,0)$. So, $M=(0,1)$ and $S=(0,\frac{3}{2}).$

An easy computation reveals that:

$$B'=(\frac{-2}{5}, \frac{16}{5}), \\ C'=(\frac{2}{5}, \frac{16}{5}).$$

Now, $P$ is the intersection of the line $y=2$ and the line $y=\frac{-4}{3}x+\frac{8}{3}$ (the line passing through $B'$, $E$, and $C$); hence:

$$P=(\frac{1}{2}, 2).$$

On the other hand, $Q$ is the intersection of the line $y=\frac{3}{2}$ and the line $y=2-x$ (the line passing through $A$ and $C$); hence:

$$Q=(\frac{1}{2}, \frac{3}{2}).$$

At this stage, it is easy to note that $|AS|=|AP|=|SQ|=\frac{1}{2}.$

We are done.

Reza Rajaei
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