# Proving that a point is a midpoint

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 $$\overline{AB}$$. 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.

• Do you allow complex numbers? If yes, the proof is pretty direct. – Calvin Lin Apr 2 at 5:26
• @Ángela Flores I solved your problem. If you want to see my solution, show please your attempts. – Michael Rozenberg Apr 2 at 5:57
• No, I cannot use complex numbers. The tools allowed are parallelism, similarity, congruences. – Ángela Flores Apr 2 at 6:21
• 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. – Ángela Flores Apr 2 at 6:27

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 $$AB$$.

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 $$|MN| = |AN|$$.)