A circle is divided into equal arcs by $n$ diamtrs. Prove the feet of the $\perp$s from a point $M$ in circle to them are vertices of a regular n-gon

Question

A circle is divided into equal arcs by $n$ diameters. Prove that the bases of the perpendiculars dropped from an arbitrary point $M$ inside the circle to these diameters are vertices of a regular n-gon.

Let the center of the circle be $O$. So, for any feet of the perpendicular $N_i$, $N_j$ and $N_k$ we have that $MN_iON_j$ and $MN_iON_k$ form a cyclic quadrilateral because opposite angles add upto $180$ degrees. But, since ther is a unique circle through $M, N_i, O$ these two circles are coincident and equal. In general, all those points lie on a quadrilateral with diameter $MO$.

I have proved that they all lie on a circle but am unable to prove that the chords or the arcs are equal. How do I proceed? Please help.

Thanks.

Answer 1

Passing from $N_i$ to $N_{i+1}$ diameter $ON_i$ and perpendicular $MN_i$ both rotate by $\pi/n$ (see picture below). It is then immediate that $\angle N_i CN_{i+1}=2\pi/n$, where $C$ is the midpoint of $OM$.

Answer 2

The circle with $OM$ as diameter inside another circle has equally spaced [at angle increments $ ( \pi/n)] $ radial rays cutting it (with some offset $\alpha$) at $N_i$

$$ ON_i = OM\, \cos ( \frac{ \pi}{n} \cdot i + \alpha ), $$

which is directly polar equation of the circle with $O$ as common center. Straightaway. Points $ ( O,N_i, M, O_j , O )$ are not labelled. Rays $M N_i$ not drawn.

Answer 3

An alternative solution is to use the fact that all points on a circular arc passing through two points $A$ and $B$ have the same angle to the segment $AB$.

In the figure above the situation for $n=3$ is shown. We know from symmetry considerations that the angles between the perpindiculars from point $M$ are $\frac{180}{n}$, in this case $60$ degrees.

Like point $M$, point $N_i$ is on the same circular arc passing through $N_k$ and $N_j$. Therefore the angle that $N_i$ has to the segment $N_kN_j$ must also be $60$ degrees.

In the same way, point $N_k$ is on the same circular arc passing through $N_i$ and $N_j$. Therefore the angle that $N_k$ has to the segment $N_iN_j$ must also be $60$ degrees.

A triangle with three $60$ degree angles is a regular n-gon.

A similar argument can be made for $n>3$.