Let $ABCD$ be a rectangle show a proof if and only if...

Question

question

Let $ABCD$ be a rectangle. We consider the points $M\in (BC)$ and $N\in CD$ such that $[BM]\equiv [DN]$ and $D\in(NC)$. Look like

a) $\angle MAN=90$ if and only if $ABCD$ is a square

b) the middle of the segment $[MN]$ is found on $BD$ if and only if $ABCD$ is a square

idea+drawings

Overall I did the first point (a) and half of the second one (b) and I got stuck on a proof.I will show you how I solved the other points, in the hope this will help you help me solve the last proof.

POINT A

As you can see I noted $\angle MAB=x $ and $\angle MAD=y$ and I showed that traingles $ADN\equiv ABM$. Knowing that $x+y=90$ I showed by some complementary angles that $\angle MAN=x+y$

I did the same thing as above, showing the congruence of the 2 triangles by some complementary angles. Showing that $ADN\equiv ABM$ I knew that $AD=AB => ABCD square$

POINT B

We show again that traiangles $ADN\equiv ABM$ which make triangle $MAN$ isosceles. This means that $AO$ is also going to be the height of the triangle, not only the median. which means that

$\angle AOM=\angle ABM=90 => ABMO $ inscribed traingle $=> \angle AMN=\angle ABO$.

From the first point we know that if $ABCD$ is square then $\angle MAN=90 (+) $ traingle $MAN$ is isosceles $=> \angle AMN=45=\angle ABO$

If $\angle ABO=45 => BO\in BD => O\in BD$

I don't know what to do when we know that $O\in BD$ and have to show that $ABCD$ is square. Hope one of you can help me! Thank you!

Answer 1

Let $|BC| = b$, $|CD|= d$, $|BM| = |DN| = x$.

$MNC$ is a right triangle, so $|OM| = |ON| = |OC|$.
From $O$, drop perpendiculars to sides $BC$, $DC$ with feet of $F$ and $G$.
Show that $F$ is the midpoint of $MC$, and $G$ is the midpoint of $CN$.

$BOF$ and $BDC$ are similar triangles, so

$$ \frac{|BF|}{|FO|} = \frac{|BC|}{|CD|} \Rightarrow \frac{ (b+x)/2 } { (c+x) / 2 } = \frac{b}{d} . $$

Hence, $b=d$ and so $ABCD$ is a square.

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Note: This solution is essentially coordinate geometry in disguise, where we calculate the slope of the diagonal in two ways.

Answer 2

Draw $NF // BC$ as shown. Extend $BD$ to cut $NF$ at $F.$

Let $BD$ cuts $MN$ at $O$ $$$$ $(i)$ Suppose that $ABCD$ is a square. We want to prove that $MO=ON$.

Proof:

Note that $\angle DFN=45^{\text o}.$

Show that $NF=BM$ and then $\Delta OBM \cong \Delta OFN.$

This implies $OM=ON$

$$$$ $(ii)$ Assume that $OM=ON$. We want to prove that $ABCD$ is a square.

It is just the reverse process of $(i)$.

Prove that $\Delta OBM \cong \Delta OFN.$

Hence $NF=BM$ which implies $NF=ND$.

This implies that $\angle DFN=45^{\text o},$ and consequently $ABCD$ is a square.