Formula lui Heron

[Andi Brojbeanu]

Aria unui triunghi ABC este data de formula
\( S=\sqrt{p(p-a)(p-b)(p-c)} \), unde \( AB=c, BC=a, AC=b \), \( 2p=a+b+c \), iar \( S=\sigma[ABC] \).

[Andi Brojbeanu]

Fie \( ABC \) un triunghi si \( [AD] \) inaltimea dusa din varful \( A \), \( D\in BC \). Notam \( h_a=AD \). Conform teoremei lui Pitagora generalizata, obtinem \( c^2=a^2+b^2-2ab\cdot cos C \) sau \( c^2=a^2+b^2-c^2+2a\cdot DC \), de unde rezulta \( DC=\frac{a^2+b^2-c^2}{2a}. \)
In triunghiul dreptunghic \( ADC \), avem \( AD^2=AC^2-DC^2 \) sau \( {h_a}^2=b^2-{(\frac{a^2+b^2-c^2}{2a})}^2 \).
\( {h_a}^2=\frac{4a^2b^2-(a^2+b^2-c^2)^2}{4a^2}=\frac{1}{4a^2}(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2). \)
\( 2ab+a^2+b^2-c^2=(a+b)^2-c^2=(a+b+c)(a+b-c)=2p\cdot 2(p-c)=4p(p-c). \)
\( 2ab-a^2-b^2+c^2=(b+c-a)(a+c-b)=2(p-a)\cdot 2(p-b)=4(p-a)(p-b). \)
Atunci \( h_a=\sqrt{\frac{1}{4a^2}}\cdot \sqrt{4\cdot 4\cdot p(p-a)(p-b)(p-c)}=\frac{1}{2a}\cdot 4\cdot \sqrt {p(p-a)(p-b)(p-c)}=\frac{2}{a}\cdot \sqrt {p(p-a)(p-b)(p-c)} \).
Folosind formula \( S=\frac{ah_a}{2} \), obtinem ca \( S=\frac{a\cdot \frac{2}{a}\sqrt {p(p-a)(p-b)(p-c)}}{2} \), rezulta \( S=\sqrt {p(p-a)(p-b)(p-c)} \)(Formula lui Heron).