$\lceil$ https://diendantoanh...bc/#entry721138 $\rfloor$
Tìm $\Delta\,\it{DEF}$ sao cho với bất kì $\Delta\,\it{ABC}$$:$
$$\it{DE}= \it{ab}\sqrt{\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{b}+ \it{c}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{C}}{\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}}}\,\,,$$
$$\it{EF}= \it{bc}\sqrt{\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{c}+ \it{a}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{A}}{\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}}}\,\,,$$
$$\it{FD}= \it{ca}\sqrt{\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{a}+ \it{b}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{B}}{\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}}\,\,.$$
