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

traffig75
2022-09-23
Answered

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

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Jaelyn Levine

Answered 2022-09-24
Author has **9** answers

Step 1

Notice that $\mathrm{\u25b3}OAM\sim \mathrm{\u25b3}DAN$, $\mathrm{\angle}OAM=\mathrm{\angle}DAN$.

So, $\mathrm{\angle}NAM={45}^{\circ}$.

Step 2

Drop a perp from N to diagonal BD.

Notice that $\mathrm{\u25b3}TMN\cong \mathrm{\u25b3}OAM$

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

Notice that $\mathrm{\u25b3}OAM\sim \mathrm{\u25b3}DAN$, $\mathrm{\angle}OAM=\mathrm{\angle}DAN$.

So, $\mathrm{\angle}NAM={45}^{\circ}$.

Step 2

Drop a perp from N to diagonal BD.

Notice that $\mathrm{\u25b3}TMN\cong \mathrm{\u25b3}OAM$

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

Ilnaus5

Answered 2022-09-25
Author has **2** answers

Step 1

$\phantom{\rule{thickmathspace}{0ex}}\mathrm{\u25b3}ANJ\phantom{\rule{thinmathspace}{0ex}}$ is an isosceles right triangle.

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 ${45}^{\circ}$.

Step 2

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

$\mathrm{tan}\left(\mathrm{\angle}AND\right)=\frac{1}{z}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(\mathrm{\angle}MNC\right)=\frac{1+z}{1-z}=-\phantom{\rule{thinmathspace}{0ex}}\frac{1+\frac{1}{z}}{1-\frac{1}{z}}=-\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{tan}\left(\frac{\pi}{4}\right)+\mathrm{tan}\left(\mathrm{\angle}AND\right)}{1-\mathrm{tan}\left(\frac{\pi}{4}\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{tan}\left(\mathrm{\angle}AND\right)}\phantom{\rule{0ex}{0ex}}=\mathrm{tan}(\pi -(\frac{\pi}{4}+\mathrm{\angle}AND))$

It follows that $\phantom{\rule{thinmathspace}{0ex}}\mathrm{\angle}MNC+\frac{\pi}{4}+\mathrm{\angle}AND=\pi \phantom{\rule{thinmathspace}{0ex}}$, and therefore $\phantom{\rule{thinmathspace}{0ex}}x=\frac{\pi}{4}\phantom{\rule{thinmathspace}{0ex}}$ regardless of z.

$\phantom{\rule{thickmathspace}{0ex}}\mathrm{\u25b3}ANJ\phantom{\rule{thinmathspace}{0ex}}$ is an isosceles right triangle.

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 ${45}^{\circ}$.

Step 2

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

$\mathrm{tan}\left(\mathrm{\angle}AND\right)=\frac{1}{z}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(\mathrm{\angle}MNC\right)=\frac{1+z}{1-z}=-\phantom{\rule{thinmathspace}{0ex}}\frac{1+\frac{1}{z}}{1-\frac{1}{z}}=-\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{tan}\left(\frac{\pi}{4}\right)+\mathrm{tan}\left(\mathrm{\angle}AND\right)}{1-\mathrm{tan}\left(\frac{\pi}{4}\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{tan}\left(\mathrm{\angle}AND\right)}\phantom{\rule{0ex}{0ex}}=\mathrm{tan}(\pi -(\frac{\pi}{4}+\mathrm{\angle}AND))$

It follows that $\phantom{\rule{thinmathspace}{0ex}}\mathrm{\angle}MNC+\frac{\pi}{4}+\mathrm{\angle}AND=\pi \phantom{\rule{thinmathspace}{0ex}}$, and therefore $\phantom{\rule{thinmathspace}{0ex}}x=\frac{\pi}{4}\phantom{\rule{thinmathspace}{0ex}}$ regardless of z.

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