Torricelli's point

For discussing Olympiad level Geometry Problems
Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

Torricelli's point

Unread post by Tahmid » Wed Jul 08, 2015 1:21 pm

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

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm
Location: Rangpur,Bangladesh

Re: Torricelli's point

Unread post by tanmoy » Wed Jul 08, 2015 8:09 pm

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

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

Re: Torricelli's point

Unread post by Tahmid » Thu Jul 09, 2015 12:19 am

how fool i am . :( i did it by ceva with a lot of manipulation .
but tanmoy, your solution is quite easy .

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