Thanic Nur Samin wrote:We use barycentric coordinates.
Let $P\equiv (p:q:r)$. Now, we know that $pq+qr+rp=0$ [The equation of circumcircle for equilateral triangles].
Now, $D\equiv (0:q:r), E\equiv (p:0:r), F\equiv (p:q:0)$.
So, the area of $\triangle DEF$ divided by the area of $\triangle ABC$ is:
$$\dfrac{1}{(p+q)(q+r)(r+p)} \times \begin{vmatrix}
0 & q & p\\
p & 0 & r\\
p & q & 0
\end{vmatrix}$$
$$=\dfrac{2pqr}{(p+q+r)(pq+qr+rp)-pqr}$$
$$=\dfrac{2pqr}{-pqr}=-2$$.
The reason of negativity is that we took signed area.
Therefore the area of $DEF$ is twice the area of $ABC$.
joydip wrote:
Proof: Let the tangent to $(ABC)$ at $A$ meet $BP$ at $J$ .Then applying pascal's theorem on hexagon $AACPBB$ we get $JF \parallel BB \parallel AC$ . So
Zawadx wrote:There's a typo in the determinant: zero for you~
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