Problem med 7-gon diagonaler

RELATIONSHIP BETWEEN DIAGONALS in the REGULAR HEPTAGON        

           There is no respected geometry book that does not have the following problem:

 

See also

(In the book "Problems of Geometry" by Tzitzeica is problem 717.) This is the solution

     Another book also presents a personal solution of the authors (some academics)

In "Solution II" above, as in the Problem 717 solution, a Theorem of PTOLEMEU is used. Without this theorem, trying only reasonings with similarities of triangles, the authors tried in "Solutia I".

     The notations in the figure above correspond to this solution. First of all, notice the congruence of the angles:

$$\angle BAC \equiv \angle CAD \equiv \angle CDB \equiv \angle ADB =\frac{\pi}{7},$$

$$\angle ABD \equiv \angle ACD =\frac{4\pi}{7}.$$

Note now that the quadrilateral $BCDN$ is rhombus ($CD \parallel BE\;,BC \parallel AD\;$and $DB$ bisect $\angle CDN$). It follows from here

$$\angle DNM = \angle DCM=\frac{4\pi}{7}\;,\;ND=BC=a.$$ 

      We have the similarity of triangles $\Delta ABM \sim \Delta ACD$ where do we get

$\frac{BM}{CD}=\frac{AM}{AD}=\frac{AB}{AC}\Leftrightarrow$

$$\frac{BM}{a}=\frac{AM}{c}=\frac{a}{b}. \tag{1}$$

 As the third angle in a triangle, we have $\angle DMN \equiv \angle DMC=\frac{2\pi}{7}$ so $\Delta DMN \equiv \Delta DMC$, so $ND=DC=a$. 

       Further $AM=AN\;$ (from angle calculations we have $\angle ANM=\pi-\angle DNM=\pi-\frac{4\pi}{7}=\frac{3\pi}{7},\;\angle AMN=\pi-2\cdot \frac{2\pi}{7}=\frac{3\pi}{7}$) so 

$AM=AN=AD-ND=c-a$. Then the second equation of (1) is written

$$\frac{c-a}{c}=\frac{a}{b}$$

which immediately results the relationship $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$.

$\blacksquare$ 

           A demonstration without words is here. It is a true demonstration without words.

          Let's reproduce the hexagon, starting with the side $CD$. Obviously we have $CD \parallel IK$.

          We observe from angle calculations that the points $G,D,K$ and $G,C,I$ are respectively collinear: $\angle GCD+\angle DCI=\frac{3\pi}{7}+\frac{4\pi}{7}=\pi$. Obviously we have the similarity indicated on the figure, hence the conclusion.

 

$\blacksquare\;\blacksquare$