## Torricelli's point

### Torricelli's point

Given a triangle $\Delta ABC$ . Let $\Delta ABF,\Delta BCD,\Delta CAE$ be equilateral triangles constructed outwards . Prove that $AD,BE,CF$ are concurrent .

### Re: Torricelli's point

Suppose,they are not concurrent. $\Delta ACF \cong \Delta ABE$.Suppose,$CF \cap BE=G$.Then $\angle FGB=\angle AGF=60 \circ$.In the same way,we can prove that $\angle DGB=60 \circ$.So,$\angle AGD=180$.$\therefore CF,BE,AD$ are concurrent.

"Questions we can't answer are far better than answers we can't question"

### Re: Torricelli's point

how fool i am . i did it by ceva with a lot of manipulation .

but tanmoy, your solution is quite easy .

but tanmoy, your solution is quite easy .