# $O$ is intersection of diagonals of the square $ABCD$. If $M$ and $N$ are midpoints of $OB$ and $CD$ respectively ,then $\angle ANM=?$

$$O$$ is intersection of diagonals of the square $$ABCD$$. If $$M$$ and $$N$$ are midpoints of the segments $$OB$$ and $$CD$$ respectively, find the value of $$\angle ANM$$.

Here is my approach:

Assuming the length of the square is $$a$$. We have $$\tan(\angle AND)=2$$ and I draw a perpendicular segment from $$M$$ to $$NC$$ and calling the intersection point $$H$$ then $$\tan (\angle MNH)=\dfrac{\frac34a}{\frac a4}=3$$ ($$MH$$ can be found by Thales Theorem in $$\triangle BDC$$) Hence

$$\angle ANM=180^{\circ}-(\tan^{-1}2+\tan^{-1}3)=45^{\circ}$$

I'm looking for other approaches to solve this problem if it is possible.

Intuitively, If I drag the point $$N$$ to $$D$$ and $$M$$ to $$O$$ (the angle is clearly $$45^{\circ}$$ here) then by moving $$N$$ from $$D$$ to $$C$$ and $$M$$ from $$O$$ to $$B$$ with constant speed, I think the angle remain $$45^{\circ}$$. But I don't know how to prove it.

• You can use the law of cosines to get $\cos x=\frac{\sqrt 2}{2}$ Oct 18, 2021 at 1:17

Notice that $$\triangle OAM \sim \triangle DAN$$, $$\angle OAM = \angle DAN$$.

So, $$\angle NAM = 45^\circ$$.

Drop a perp from $$N$$ to diagonal $$BD$$.

Notice that $$\triangle TMN \cong \triangle OAM$$

So, $$AM = MN$$ leading to $$\angle ANM = 45^\circ$$.

• Alternative ending: from similarity we obtain $\angle ADM=\angle ANM$ and hence, $AMND$ is a cyclic quadrilateral. Thus, $\angle ANM=\angle ADM=45^\circ$. Oct 19, 2021 at 5:57

Let a side of the the square be 1 unit.

Let the circle ADN cut BD at Z and AC at Z'.

Note that DN = NC; $$AC = BD = \sqrt 2$$ and by symmetry, BZ = CZ' = x, say.

By power of a point, $$x \times \sqrt 2 = 1 \times 0.5$$. This means $$x = \dfrac {\sqrt 2}{4}$$.

That is Z is your M and hence, $$\angle ANM = \angle ADB = 45^0$$

Not sure how rigorous it is, but here's an alternate "dragging the point" explanation using complex numbers:

Let $$\mathbf{a}=(1,0)$$ and $$\mathbf{b}=(0,-1)$$ be unit vectors in the complex plane. As shown in Figure 1, letting $$\mathbf{c}=1/2(\mathbf{a}+\mathbf{b})$$, then, by the paralellogram law for addition, $$\angle \mathbf{0}\mathbf{b}\mathbf{c}=\pi/4$$.

For $$0\leq t\leq 1$$, consider the lines $$\mathscr{L}(t)=Re^{\theta i}\mathbf{b}=Re^{\left(\theta+\frac{3\pi}{2}\right)i}$$ and $$\mathscr{M}(t)=Re^{\theta i}\mathbf{a}=i\mathscr{L}(t),$$ where $$R=\sqrt{1+t^2}$$ and $$\theta=\arcsin\left(\frac{t}{\sqrt{1+t^2}}\right).$$ The lines $$\mathscr{L}$$ and $$\mathscr{M}$$ are perpendicular, $$|\mathscr{L}(t)|=|\mathscr{M}(t)|$$, and $$\mathscr{N}(t)=\frac{1}{2}\big(\mathscr{L}(t)+\mathscr{M}(t)\big)=\frac{1}{2}\big((1+i)\mathscr{L}(t)\big)=\frac{\sqrt{2}}{2}e^{\frac{\pi i}{4}}\mathscr{L}(t)$$ is also a line. We need only show that $$\mathscr{N}$$ is the diagonal line between $$\mathbf{c}$$ and $$\mathbf{a}$$ and we're done. This follows from the fact that $$\mathscr{L}(1)=1-i$$ and $$\mathscr{M}(1)=1+i$$, so that $$\mathscr{N}(1)=\mathbf{a}$$. Hence, for all $$t$$, $$\angle\mathbf{0}\mathscr{\mathscr{L}(t)}\mathscr{\mathscr{N}(t)}=\frac{\pi}{4}. \qquad \square$$

$$\tag{Fig. 1}$$

Hint: $$\;\triangle ANJ\,$$ is an isosceles right triangle.

[ EDIT ] The following is about the second part of OP's question.

Intuitively, If I drag the point $$N$$ to $$D$$ and $$M$$ to $$O$$  [...]  then by moving $$N$$ from $$D$$ to $$C$$ and $$M$$ from $$O$$ to $$B$$ with constant speed, I think the angle remain $$45^{\circ}$$.

Let $$\,AB=1\,$$, $$\,DN=z \in \left[0, \frac{1}{2}\right]\,$$ and $$\,\frac{OM}{OB}=\frac{DN}{DC}\,$$ so that $$\,z=\frac{1}{2}\,$$ corresponds to the original diagram and $$\,z=0\,$$ corresponds to $$\,M \mapsto O, N \mapsto D\,$$. Then $$\,OM = \frac{OB}{DC} \,DN = \frac{z}{\sqrt{2}}$$, and:

$$\tan\left(\angle AND\right) = \frac{1}{z} \\ \tan\left(\angle MNC\right)=\frac{1+z}{1-z} = -\,\frac{1+\frac{1}{z}}{1 - \frac{1}{z}} = -\,\frac{\tan\left(\frac{\pi}{4}\right)+\tan\left(\angle AND\right)}{1-\tan\left(\frac{\pi}{4}\right)\,\tan\left(\angle AND\right)} \\ = \tan\left(\pi -\left(\frac{\pi}{4}+\angle AND\right)\right)$$

It follows that $$\,\angle MNC + \frac{\pi}{4} + \angle AND = \pi\,$$, and therefore $$\,x = \frac{\pi}{4}\,$$ regardless of $$\,z\,$$.

• Thanks! I really like your answer too, but unfortunately I can't check mark more than one answer. Oct 19, 2021 at 9:43

Note that $$\small \triangle ADN$$ is a right triangle with perpendicular sides in the ratio $$1:2$$.

Draw $$\small NT$$ parallel to $$\small CA$$.

From $$\small \triangle DOC$$, $$\small DT=TO$$ (midpoint theorem).

Thus, can you see $$\small TM=2\ TN$$?

Therefore $$\small \triangle MTN$$ is also a right triangle with perpendicular sides in the ratio $$1:2$$.

$$\small \therefore\triangle ADN\sim\triangle MTN\implies\angle DAN=\angle TMN\implies ADNM\text{ is cyclic}$$

$$\small \implies x= \angle ANM=\angle ADM=45^\circ$$

If you rotate the square $$ABCD$$ clock-wise $$90^{\circ}$$ around the center $$O$$ of the square $$ABCD$$, then vertex $$B$$ is rotated to vertex $$D$$, which means that the segment $$OB$$ is rotated to the segment $$OD$$, which means that the midpoint $$M$$ of $$OB$$ is mapped to the midpoint $$K$$ of $$OD$$.

Consequently, since under the rotation, the vertex $$A$$ is rotated to the vertex $$B$$, the segment $$AM$$ is rotated to the segment $$BK$$, which means that $$AM = BK$$ and $$AM \, \perp\, BK$$.

Furthermore, $$KN$$ is a midsegment in the triangle $$COD$$, so $$KN = \frac{1}{2}\,OC$$ and $$KN \, || \, OC$$. However, $$BM = \frac{1}{2}\,OB = \frac{1}{2}\,OC = KN \,\, \text{ and } \,\, KN \, || \, BM$$ which means that $$BMNK$$ is a parallelogram and thus $$MN = BK \,\, \text{ and } \,\, MN \, || \, BK$$ Therefore, $$MN = BK = AM \,\, \text{ and }\,\, MN \, \perp \, AM$$ Consequently, $$\angle\, AMN = 90^{\circ} \,\, \text{ and } \,\, AM = MN$$ which means that the triangle $$AMN$$ is equilateral right-angled triangle, so $$\angle\, ANM = \angle \, NAM = 45^{\circ}$$