geometrytriangles
Question –
equilateral triangles DBC, ECA and FAB are constructed externally on the sides BC,CA,AB respectively of triangle ABC. Prove that AD, BE, and CF are also concurrent.
My attempt –
I know this is fermat point and also that this question is already been answered but I am not getting how to proof this using ceva theorem to prove concurrency….I know the proof using rotation… kindly help me thanks …
Let $\Delta ABC$ be a triangle, denote by $M,N,P$ the mid points of the sides opposite to $A,B,C$. Let $D$ be a point on $BC$. The parallel to $AD$ through $M$ intersects $NP$ in a point denoted by $D'$.
Construct $E',F'$ in a similar way on the sides of $\Delta MNP$.
Because $MN$, $ND'$, $D'M$ are parallel to (respectively) $AB$, $BD$, $DA$, we obtain a triangle similarity, so $D'N:DB=MN:AB=1:2$. This gives $D'N:D'P=DB:DC$.
We finally apply Ceva (and the reciprocal) in the two triangles $\Delta ABC$, $\Delta MNP$: $$ \frac{D'N}{D'P}\cdot \frac{E'P}{E'M}\cdot \frac{F'M}{F'N} = \frac{DB}{DC}\cdot \frac{EC}{EA}\cdot \frac{FA}{FB} = -1\ . $$ $\square$
Later edit: Quick picture inserted:
Here is another approach to the problem. I will outline the steps involved.
Extend $MF$ to $G$ s.t. $MF=FG$.
Join the lines as shown.
Prove that
(a) $BG=MC$, $\angle GBD= \angle MCD$, $BD=CD$ and hence $\Delta GBD \cong \Delta MCD.$
(b) Use the fact that $\Delta GBD \cong \Delta MCD$ to prove that $\Delta DMG$ is equilateral.
(c) Result follows because $\Delta FDM$ is just half of $\Delta DMG.$
Best Answer
Let $A'$ be the intersection of $AD$ and $BC$. Then (if all angles are less than $2\pi/3$):
$$\frac{A'B}{A'C} = - \frac{\text{Area}(ABD)}{\text{Area}(ACD)} = - \frac{AB \cdot BD \cdot \sin{(B + \pi/3})}{AC\cdot CD \cdot \sin{(C + \pi/3)}} = - \frac{AB \sin{(B + \pi/3)}}{AC\sin{(C + \pi/3)}}.$$
Write the other two similar ratios and multiply to get -1. The equality between the first and the last term remains valid for angles exceeding $2\pi/3$.