Im gleichseitigen Dreieck gilt mit Pythagoras:

$h_a=\sqrt{a^2-{\left({\displaystyle \frac{a}{2}}\right)}^2}=\sqrt{{\displaystyle \frac{3\cdot a^2}{4}}}={\displaystyle \frac{\sqrt{3}}{2}}ah\_a=\backslash sqrt\{a^2-\backslash left(\backslash dfrac\{a\}\{2\}\backslash right)^2\}=\backslash sqrt\{\backslash dfrac\{3\backslash cdot\; a^2\}\{4\}\}=\backslash dfrac\{\backslash sqrt\{3\}\}\{2\}a$

$A_{{\mathrm{\Delta }}_G}={\displaystyle \frac{1}{2}}\cdot g\cdot h={\displaystyle \frac{1}{2}}\cdot a\cdot {\displaystyle \frac{\sqrt{3}}{2}}a={\displaystyle \frac{\sqrt{3}}{4}}\cdot a^{2}A\_\{\backslash Delta\_G\}=\backslash dfrac\{1\}\{2\}\backslash cdot\; g\backslash cdot\; h=\backslash dfrac\{1\}\{2\}\backslash cdot\; a\backslash cdot\; \backslash dfrac\{\backslash sqrt\{3\}\}\{2\}a=\backslash dfrac\{\backslash sqrt\{3\}\}\{4\}\backslash cdot\; a^2$

Betrachte das Dreieck $ADSADS$.

Die Seite $ADAD$ ist die Höhe des gleichseitigen Dreiecks $ABCABC$.

Die Seitenhalbierenden (Höhen) schneiden sich im Verhältnis $2:12:1$.

Für die Strecke $pp$ gilt dann:

$p={\displaystyle \frac{2}{3}}\cdot \overline{AD}={\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{\sqrt{3}}{2}}a={\displaystyle \frac{\sqrt{3}}{3}}ap=\backslash dfrac\{2\}\{3\}\backslash cdot\backslash overline\{AD\}=\backslash dfrac\{2\}\{3\}\backslash cdot\backslash dfrac\{\backslash sqrt\{3\}\}\{2\}a\; =\backslash dfrac\{\backslash sqrt\{3\}\}\{3\}a$

Für $qq$ gilt: $q={\displaystyle \frac{1}{3}}\cdot \overline{AD}={\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{\sqrt{3}}{2}}a={\displaystyle \frac{\sqrt{3}}{6}}aq=\backslash dfrac\{1\}\{3\}\backslash cdot\backslash overline\{AD\}=\backslash dfrac\{1\}\{3\}\backslash cdot\backslash dfrac\{\backslash sqrt\{3\}\}\{2\}a\; =\backslash dfrac\{\backslash sqrt\{3\}\}\{6\}a$

Im rechtwinkligen Dreieck $AHSAHS$ ist

$h_P=2ah\_P\; =\; 2a$. Dann gilt wieder mit Pythagoras:

$k^2=p^2+h_P^2={\left({\displaystyle \frac{\sqrt{3}}{3}}a\right)}^2+(2a{)}^{2}k^2=p^2+h\_P^2=\backslash left(\backslash dfrac\{\backslash sqrt\{3\}\}\{3\}a\backslash right)^2+(2a)^2$

$k^2={\displaystyle \frac{1}{3}}{a}^2+4a^2={\displaystyle \frac{13}{3}}{a}^{2}k^2=\backslash dfrac\{1\}\{3\}a^2+4a^2=\backslash dfrac\{13\}\{3\}a^2$

$\Rightarrow k=\sqrt{{\displaystyle \frac{13}{3}}}a={\displaystyle \frac{\sqrt{39}}{3}}a\backslash Rightarrow\backslash ;k=\backslash sqrt\{\backslash dfrac\{13\}\{3\}\}a=\backslash dfrac\{\backslash sqrt\{39\}\}\{3\; \}a$

Die Seitenkantenlänge $kk$ ist:

Im rechtwinkligen Dreieck $HDSHDS$ gilt:

$h_{\mathrm{\Delta }}^2=h_P^2+q^2=(2a{)}^2+{\left({\displaystyle \frac{\sqrt{3}}{6}}a\right)}^2=4a^2+{\displaystyle \frac{1}{12}}{a}^2={\displaystyle \frac{49}{12}}{a}^{2}h\_\{\backslash Delta\}^2=h\_P^2+q^2=(2a)^2+\backslash left(\backslash dfrac\{\backslash sqrt\{3\}\}\{6\}a\backslash right)^2=4a^2+\backslash dfrac\{1\}\{12\}a^2=\backslash dfrac\{49\}\{12\}a^2$

$\Rightarrow h_{\mathrm{\Delta }}={\displaystyle \frac{\sqrt{49}}{\sqrt{12}}}a={\displaystyle \frac{7}{2\sqrt{3}}}a={\displaystyle \frac{7\cdot \sqrt{3}}{6}}a\backslash Rightarrow\backslash ;h\_\{\backslash Delta\}=\backslash dfrac\{\backslash sqrt\{49\}\}\{\backslash sqrt\{12\}\}a=\backslash dfrac\{7\}\{2\backslash sqrt\{3\}\}a=\backslash dfrac\{7\backslash cdot\; \backslash sqrt\{3\}\}\{6\}a$