FERMAT POINT

ABC is a Triangle and P be a Fermat Point Inside it.draw three equilateral triangle based on the three sides i.e$\triangle ABA'$, $\triangle ACC'$, $\triangle BCB'$ respectively.Join $AB'$,$BC'$ and$CA'$ .Show that $ AB'$,$BC'$ and $CA'$ pass through a single piont i.e they are concurrent.

Rotation:

ABC is a Triangle . Let P Be any point join $AP,BP$ and $CP$. Now if we rotate the $\triangle ABP$ about the point at B $ 60 ^{\circ} $ anti clockwise we will get $\triangle BP'A'$.

SHORTEST DISTANCE:

Join the point P and P'.Now In the triangle BPP' we have

BP-BP'

$\angle PBP'=60 ^{\circ} $, SO $\triangle BPP' $ is a equilateral triangle. so $BP=BP'=PP'$

and also $AP'=AP$ (Length remain unchange after Rotation).

So from the point $A'$ to $C$ the path is $A'P'+PP'+PC$.This path will be Shortest distance if $A'P'+PP'+PC$ i.e A'C be a straight line. and also $AP+PB+AB=A'P'+PP'+PC$

the shortest path betwween two points is a straight line and so $ PA+PB+PC$ reaches its minimum if and only if the point $p$ and $P'$ lie on the line $A'C$

By symmetry it follows that $ P $ must also lie on the line $BC'$ and $AB'$.

So the point of intersection of these lines is a fermat point of a $\triangle ABC$.

EQUILATERAL TRIANGLE :

Now the triangle $AA'B$ we have

$A'B=AB$ (length remain unchange due to rotation)

$\angle A'BA =60^{\circ}$. so the triangle $AA'B $ is a equilateral triangle .

similarly for the other two triangles $AC'C$ and $BB'C$

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