In the square $ABCD$ on the sides $AD$ and $DC$, the points $M$ and $N$ are selected so that $\angle BMA = \angle NMD = 60 { } ^ \circ $. Find the value of the angle $MBN$.
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Tags: geometry, angles, square, Angle Chasing, equal angles
vanstraelen
06.08.2020 22:56
$\angle MBN = 45^{\circ} $.
yofro
07.08.2020 01:57
@above how? I am getting $30^{\circ}$. $\angle ABM = 30$, $\angle DNM=30$, $\angle BNC=60$, $\angle NBC=30$, $\angle MBN=\boxed{30^{\circ}}$. I may be wrong when saying that $\angle BNC=60$, I assumed $\triangle MNB$ is right.
yofro
07.08.2020 02:01
WLOG let $M$ be the midpoint of $AD$, and let $N$ be $C$. We have $\angle BMN=60$, $\angle ABM=30$, $\boxed{\angle MBN=\boxed{60^{\circ}}}$
Inconsistent
07.08.2020 02:15
@above Both your results are wrong. Why is $MN \perp NB$? Also, you can't WLOG $M$ to be the midpoint of $AD$ or $N$ to be $C$, it's already defined. And somehow this time you got 60 degrees?
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yofro
07.08.2020 03:59
Yah, you're probably right. How do you get 45 then?
ApraTrip
07.08.2020 04:03
I mean, you could kill it with law of cosines.
SilverDragon246
07.08.2020 05:49
WLOG let $AM = 1$ then through a series of $30-60-90$ triangles and using the fact that $AB = AD$ we get that $MB = 2$ and $MN = 2\sqrt{3} - 2$. We also know that $\angle BMN = 60^{\circ}$.
Let $x = BN$. Using law of cosines on triangle $BMN$ we get that $x^2 = 24 - 12\sqrt{3}$.
Now by law of sines on triangle $BMN$, if we let $\angle MBN = B$, we get that
$\dfrac{\dfrac{\sqrt{3}}{2}}{\sqrt{28-12\sqrt{3}}} = \dfrac{\sin(B)}{2\sqrt{3} - 2}$
Squaring both sides and simplifying you get that $\sin^2(B) = \dfrac{1}{2}$ which implies that $\angle MBN = 45^{\circ}$.
rafaello
08.08.2020 10:18
We see that since $\angle AMB=\angle BMN$ and since $\angle MDB=\angle BDN=45^{\circ}$, thus $\angle MNB=\angle BNC$, because $B$ is excircle centre. Now considering triangle $DMN$, we compute that $\angle DNM=30^{\circ}$, thus $\angle MNB=\angle BNC=75^{\circ}$. Considering triangle $BMN$ we compute that $\angle BMN=45^{\circ}$. $\boxed{\text{Answer. } 45^{\circ}.}$
zuss77
08.08.2020 13:18
$DB, MB$ - internal and external angle bisectors of $\triangle DMN$, so $B$ - its $D$-excenter. So $\angle MBN=90^\circ-\angle D/2=45^\circ$.
