# Square with midpoints of $AD$ and $CD$

Square $$ABCD$$ is given; $$MA=MD=ND=NC$$. Show $$AF=AB$$. 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?

• $\angle BFM=\angle CFN=180^\circ-\angle MCD-\angle CNB=180^\circ-\angle MCD -\angle CMD=90^\circ$.
– user678090
Aug 20 '19 at 10:53
• For this problem, I think the analytic method will my first choice.
– user678090
Aug 20 '19 at 10:57

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 $$

So $$ABF$$ is isosceles.

• I can not believe. Nobody can see simplitines of this aproach.
– Aqua
Aug 20 '19 at 11:13
• If I remember, $B \mapsto C$ because $SB = SC$ and $\angle BSC = 90 ^\circ$, right? Aug 20 '19 at 11:23
• Yes, that is correct @AndrewRogers
– Aqua
Aug 20 '19 at 11:24
• 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. Aug 20 '19 at 11:26
• This is a theorem you could easly prove it your self.
– Aqua
Aug 20 '19 at 11:29

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

• Yes, this "like mad" approach creates plenty of congruent triangles which then add up to what we want! Aug 20 '19 at 10:59
• Thank you for your response! Can you see the edit to tell me if I can continue my idea? Aug 20 '19 at 12:23
• @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.
– user694818
Aug 20 '19 at 12:36

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!

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

Drop perpendicular $$AG$$ and denote $$\angle ABG=x$$:

$$\hspace{3cm}$$ 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 $$AF=AB$$.

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