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Square $ABCD$ is given; $MA=MD=ND=NC$. Show $AF=AB$. enter image description here

The first thing I noticed was $\triangle CDM \cong \triangle BCN$ and we obtain $CM = BN$ and $\angle MCD = \angle NBC$. Now I am trying to show $\angle BFM = 90 ^\circ$ but I am having difficulties with this and don't know if it will help with the solution. Would appreciate help of any kind!

Edit: Thank you for your help! Now I see how $\angle BFM = 90^\circ$. $\angle BAM + \angle BFM = 180 ^\circ$, thus $ABFM$ is a cyclic quadrilateral. Does this help?

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    $\begingroup$ $\angle BFM=\angle CFN=180^\circ-\angle MCD-\angle CNB=180^\circ-\angle MCD -\angle CMD=90^\circ$. $\endgroup$
    – user678090
    Aug 20 '19 at 10:53
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    $\begingroup$ For this problem, I think the analytic method will my first choice. $\endgroup$
    – user678090
    Aug 20 '19 at 10:57
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Let $S$ be a center of a square.

Notice that rotation around $S$ for $90^{\circ}$ takes $BN$ to $CM$ ($B\mapsto C$, $C\mapsto D$ ... and since $DM = CN$ we have $N\mapsto M$) so $\angle BFM = 90^{\circ}$.

Now we see that $ABFM$ is cyclic, so $$<AFB = <AMB = <DMC = 90-<DCM = 90-<CBN = < ABF$$

So $ABF$ is isosceles.

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  • $\begingroup$ I can not believe. Nobody can see simplitines of this aproach. $\endgroup$
    – Aqua
    Aug 20 '19 at 11:13
  • $\begingroup$ If I remember, $B \mapsto C$ because $SB = SC$ and $\angle BSC = 90 ^\circ$, right? $\endgroup$ Aug 20 '19 at 11:23
  • $\begingroup$ Yes, that is correct @AndrewRogers $\endgroup$
    – Aqua
    Aug 20 '19 at 11:24
  • $\begingroup$ And the angle between $BN$ and $CM$ is equal to the angle of rotation. Is there a theorem about this because I haven't read this in my book? I know it from the problems I've solved. $\endgroup$ Aug 20 '19 at 11:26
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    $\begingroup$ This is a theorem you could easly prove it your self. $\endgroup$
    – Aqua
    Aug 20 '19 at 11:29
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Yes, they are perpendicular. $\angle BNC \cong \angle CMD$ by CPCTC, and therefore $\triangle CFN \sim \triangle CDM$ and thus $\angle CFN$ is a right angle.

Moving on from there, let's just draw parallels to BM and CN through A, B, C, and D.

enter image description here

From this, it should be pretty clear that $AB=AF$.

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  • $\begingroup$ Yes, this "like mad" approach creates plenty of congruent triangles which then add up to what we want! $\endgroup$ Aug 20 '19 at 10:59
  • $\begingroup$ Thank you for your response! Can you see the edit to tell me if I can continue my idea? $\endgroup$ Aug 20 '19 at 12:23
  • $\begingroup$ @Andrew Rogers It's not clear to me how a cyclic quadrilateral will help. You could use Ptolemy's Theorem if you knew FM and FB, but I don't know how those would be achieved. $\endgroup$
    – user694818
    Aug 20 '19 at 12:36
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Extend $FM$ and $AB$ to meet at $K$.

Thus, since $\measuredangle KFB=90^{\circ}$ and $FA$ is a median of $\Delta KFB,$ we obtain: $$KA=AB=FA$$ and we are done!

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  • $\begingroup$ Well done! :) It looks so simple. $\endgroup$ Aug 20 '19 at 16:52
  • $\begingroup$ @Andrew Rogers Nice problem! $\endgroup$ Aug 20 '19 at 17:07
  • $\begingroup$ @Andrew Rogers See the farruhota's picture. It must help. $\endgroup$ Aug 20 '19 at 17:18
  • $\begingroup$ How do we know $A$ is the midpoint of $KB$? $\endgroup$ Aug 20 '19 at 17:26
  • $\begingroup$ Oh, $\triangle AMK \cong CDM$. $\endgroup$ Aug 20 '19 at 17:29
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Drop perpendicular $AG$ and denote $\angle ABG=x$:

$\hspace{3cm}$enter image description here

Note: $$x=\angle ABG=\angle BNC \quad \text{(because $AB||CD$})\\ \angle CFN=90^\circ \quad \text{(because $\Delta CFN\sim \Delta CDM$)} \\ \Delta ABG\cong \Delta BCF \quad \text{(because correspondig angles and one side are equal)}\\ BF=2CF=4FN \quad \text{(because $\tan x=2$)}\\ FG=BF-BG=2CF-BG=2BG-BG=BG.$$ So, the perpendicular $AG$ is also median, hence $\Delta AFB$ is an isosceles triangle and $AF=AB$.

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  • $\begingroup$ Thank you! What is $\tan$? $\endgroup$ Aug 20 '19 at 12:30
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    $\begingroup$ $\tan x=\tan \angle CMD=\frac{CD}{DM}=2$ $\endgroup$
    – farruhota
    Aug 20 '19 at 12:31

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