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I'm trying to solve this using elementary geometry.

Let $\triangle ABC$ be a triangle and consider the squares $CBFG$ and $ACDE$. Draw $\overline{EF}$ and let $M$ be its midpoint. Draw the perpendicular to $\overline{AB}$ passing through $M$. Let $N$ be the intersection point of this perpendicular and $\overline{AB}$ . Prove that $N$ is the midpoint of $\overline{AB}$.

image of the geometry problem

Any ideas?

I've tried extending $MA$ to $A'$, with $MA=MA'$, same with $MB$. I draw $A'F$, $B'E$, expecting them to be congruent to $ABC$, but I couldn't prove it (they should be congruent). A lot of lines here and there, but I couldn't say anything relevant.

I cannot use complex numbers. The tools allowed are parallelism, similarity, congruences.

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  • $\begingroup$ Do you allow complex numbers? If yes, the proof is pretty direct. $\endgroup$ – Calvin Lin Apr 2 at 5:26
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    $\begingroup$ @Ángela Flores I solved your problem. If you want to see my solution, show please your attempts. $\endgroup$ – Michael Rozenberg Apr 2 at 5:57
  • $\begingroup$ No, I cannot use complex numbers. The tools allowed are parallelism, similarity, congruences. $\endgroup$ – Ángela Flores Apr 2 at 6:21
  • $\begingroup$ I've tried extending MA to A', with MA=MA´, same with MB. I draw A´F, B'E, expecting them to be congruent to ABC, but I couldn't prove it (they should be congruent). A lot of lines here and there, but I couldn't say anything relevant. $\endgroup$ – Ángela Flores Apr 2 at 6:27
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Let $P$ be the center of the square $ACDE$ and $Q$ be the center of the square $BCGF$.

Then $MP$ is a mid-segment of triangle $CEF$ and is therefore parallel to $CF$ and half of it. As $Q$ is the midpoint of $CF$, then $MP = CQ$. But $Q$ is the center of the square, so $CQ = BQ$. Hence $MP = BQ$ and $MP \, || \, CQ$.

Analogously, $MQ$ is a mid-segment of triangle $CEF$ and is therefore parallel to $CE$ and half of it. As $P$ is the midpoint of $CE$, then $MQ = CP$. But $P$ is the center of the square, so $CP = AP$. Hence $MQ = AP$ and $MQ \, || \, CP$.

$MPCQ$ has $MP \, || \, CQ$ and $MQ \, || \, CP$, so it is a parallelogram. Thus, $\angle \, CPM = \angle \, CQM$. Therefore $$\angle \, APM = 90^{\circ} - \angle \, CPM = 90^{\circ} - \angle \, CQM = \angle \, BQM$$

Now, due to the fact that $MP = BQ, \,\, MQ = AP$ and $\angle \, APM = \angle \, BQM$, the triangles $BQM$ and $APM$ are congruent, which yields that $$AM = BM$$ Consequently, the triangle $ABM$ is isosceles and $MN$ is its height. Hence it is also the orthogonal bisector of $AB$, i.e. $N$ is the midpoint of $AB$.

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Here is a proof by complex numbers. You can also use Coordinate Geometry in it's place, you just have to keep track of the x and y coordinates.

Let $C$ be the origin.
Let $ A = a, B = b$.
Let $N = \frac{ a+b}{2}$ be the midpoint of $AB$.
Then $ E = a - ia = (1-i) a$, $F = b + ib = (1+i)b$, $M = \frac{(1-i) a + (1+i) b } { 2} $.
Observe that $MN = \frac{ - ia + ib } { 2 } $ and $AN = \frac{ -a + b } { 2}$.

Hence, these vectors are perpendicular to each other.
In fact, we have $ |MN| = |AN|$.


Here's a proof via coordinate geometry.

Let $ A = (0,0), B = (b, 0), C = (x, y)$.
Then $ E = (-y, x), F = (b+y, b - x)$, $M = (\frac{b}{2}, \frac{b}{2})$.
Thus $M$ lies on the perpendicular bisector of $ AB$.
(Likewise, we get $|MN| = |AN|$.)

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