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Sa se arate ca (notatii standard) exista relatia

\( [ABC]=\frac {a-b}{\frac {1}{r_b}-\frac {1}{r_a}}=\frac {r^2(r_a-r_b)(r_a-r_c)(r_b-r_c)}{(a-b)(a-c)(b-c)}\ . \)

Indicatie. \( S=pr=(p-a)r_a=(p-b)r_b=(p-c)r_c \) etc.

Mathesis, pag. 254

Prima egalitate este putin gresita. Una din diferentele de la numarator sau de la numitor trebuie scrisa invers.

\( \displaystyle\frac{a-b}{\frac{1}{r_b}-\frac{1}{r_a}}=\frac{a-b}{\frac{p-b}{S}-\frac{p-a}{S}}=\frac{a-b}{\frac{a-b}{S}}=S \)

Pentru cea de-a doua:

\( \displaystyle\frac{r^2(r_a-r_b)(r_a-r_c)(r_b-r_c)}{(a-b)(a-c)(b-c)}= \)
\( \displaystyle=\frac{\frac{S^2}{p^2}\left(\frac{S}{p-a}-\frac{S}{p-b}\right)\left(\frac{S}{p-a}-\frac{S}{p-c}\right)\left(\frac{S}{p-b}-\frac{S}{p-c}\right)}{(a-b)(a-c)(b-c)}= \)
\( \displaystyle=\frac{\frac{S^2}{p^2}\cdot\frac{S(a-b)}{(p-a)(p-b)}\cdot\frac{S(a-c)}{(p-a)(p-c)}\cdot\frac{S(b-c)}{(p-b)(p-c)}}{(a-b)(a-c)(b-c)}= \)
\( \displaystyle=\frac{S^5}{S^4}=S \)