A cyclic octagon from a book named "Plane Geometry Problems"

Question

Consider a cyclic octagon $ABCDEFGH$, such that $AB=BC, CD=DE, EF=FG, GH=HA$.

Diagonals $AD, BE$ meet at $P$, and diagonals $EH, FA$ meet at point $Q$. Prove that if $M$ is the midpoint of $PQ$, then $M$ lies on segment $BF$.

I tried angles, they didn't help and first I thought there are parallel lines on the figure, but later I realised that is not true. Please help, how can I prove it?

Answer 1

Lemma If $A',B',C'$ are the midpoints of arcs $BC,CA$ and $AB$ on the circum circle for a triangle $ABC$ (wich doesn't contain $A,B,C$ respectively) then an incenter $I$ of a triangle $ABC$ is an orthocenter of a triangle $A'B'C'$. So $$ \vec{I} = \vec{A'}+\vec{B'}+\vec{C'}$$ where the circumcenter is origin of a position vectors. $\diamondsuit $

Let $X$ be a midpoint of arc $AE$ which contains $B$ and $Y$ the other midpoint of arc $AE$. Since $P$ is incenter for the triangle $ACE$ we have, by Lemma: $$ \vec{P} = \vec{B}+\vec{D}+\vec{Y}$$
and similary for $Q$: $$ \vec{Q} = \vec{F}+\vec{H}+\vec{X}$$ so, since $ \vec{Y} = -\vec{X}$ we have:
$$ \vec{M} = {1\over 2}(\vec{P}+\vec{Q}) = {1\over 2}(\vec{B}+\vec{D} +\vec{F}+\vec{H})$$

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On the other hand we see that $$arc HF +arc BD = \pi r \;\;\Longrightarrow \;\; \angle HOF +\angle BOD = \pi \;\;\Longrightarrow \;\; \angle HDF +\angle BFD = \pi/2$$ so $BF\bot DH$. Let $M'$ be intersection point of a lines $BF$ and $HD$. If we reflect $H$ across line $BF$ we get $H'$ which is orthocenter of a triangle $BDF$ so $$ \vec{H'} = \vec{B}+\vec{D}+\vec{F}$$ and $$ \vec{M'} = {1\over 2}(\vec{H}+\vec{H'}) = {1\over 2}(\vec{H} + \vec{F}+\vec{B}+\vec{D}) = \vec{M}$$

and we are done.

Answer 2

HINT.-With rectangular coordinates and without theorems.

Making line $OA$ as x-axis in rectangular coordinates, $R$ being the radius of the circle, one has $$A=(R,0)\space \space B=(r\cos(\alpha),R\sin(\alpha)\space\space C=(R\cos(2\alpha),R\sin(2\alpha))\\ D=(R\cos(2\alpha+\beta), R\sin(2\alpha+\beta))\space \space E=(R\cos(2\alpha+2\beta),R\sin(2\alpha+2\beta))\\F=(R\cos(2\alpha+2\beta+\gamma),R\sin(2\alpha+2\beta+\gamma))\\ G=(R\cos(2\alpha+2\beta+2\gamma),R\sin(2\alpha+2\beta+2\gamma))\\H=(-R\cos(\alpha+\beta+\gamma),R\sin(\alpha+\beta+\gamma))$$

The angles $\alpha,\beta,\gamma$ are arbitrary under the restriccion $\alpha+\beta+\gamma\lt 2\pi$ but the fourth angle, say $2\delta$, is determined by $2\delta=2\pi-2(\alpha+\beta+\gamma)$. This way one determine the vertices of a cyclic octagon. The calculation to finish is as follow:

►lines $AD$ and $BE$ determine the point $P$.

►lines $EH$ and $AF$ determine the point $Q$

►the midpoint of segment $\overline{PQ}$ must belong to the $BF$ line.