circleseuclidean-geometrygeometry
For reference:
Since the regular polygon ABCDE, calculate the number of sides
knowing that AC and BE forms an angle whose measurement is $135^{\circ}$.
My progress:
I found that there is an isosceles triangle
I drew some auxiliary lines but I can't find the relationship
by geogebra:
Build a parallelogram $ABDE$. Then $$\angle DEA = \angle ABD = 135^\circ = \angle DCA,$$ hence $ADEC$ is cyclic. Moreover, $\angle EAD = \angle BDA = x$, hence $$\angle CAE = \angle CAD - \angle EAD = 2x-x=x.$$ As a consequence, $CE=DE$ because the angles subtended by arcs $CE$, $DE$ of the red circle are equal.
Now, it is easy to calculate that angle $CBD$ equals $90^\circ + x$ and that the concave angle $CED$ equals $180^\circ + 2x$. Since $CE=DE$, it follows that $B$ lies on the circle with center $E$ and radius $CE=DE$. Hence $\angle DEB = 2\angle DCB = 90^\circ$. Since $DE=BE$, the right triangle $DEB$ is isosceles.
Let $F$ be the midpoint of $BE$. Let $G$ be the midpoint of $BD$ and $H$ be the midpoint of $BG$. Easy to see that triangles $GFB$ and $GFH$ are isosceles right triangles, hence $HF = HB = \frac 12BG = \frac 14 BD$, so $HD = BD - BH = 4HF - HF = 3HF$. This yields $\tan x = \frac{HF}{HD} = \frac 13$ so the answer is $$x = \arctan \frac 13.$$
Due to vector symmetry for any regular n- sided polygon average of sum of projections is from the center of the polygon:
$$ \bar p =\dfrac{\Sigma p}{n} =\dfrac{18}{6} =3.$$
Proof : Vector sum of sides =0, the above is a general relation.
Best Answer
$$45^{\circ}=\measuredangle EMC=\measuredangle EBC+\measuredangle ACB=\frac{1}{2}\left(\widehat{EC}+\widehat{AB}\right)=\frac{3}{2}\widehat{AB}=\frac{3}{2}\cdot\frac{360^{\circ}}{n}$$