[asy][asy] size(10cm); pair A = dir(70); pair B = dir(210); pair C = dir(330); draw(A--B--C--A--cycle); pair D = 2*B-C; draw(A--D--B); pair F = 2*A-C; pair E = 2*C-A; draw(D--E--F--D--cycle); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); [/asy][/asy] It is clear that $p(ABC)<p(ADE)$, thus $2p(ABC)=p(ADE)$. Let $F$ be the reflection of $C$ across $A$. Then $\angle ADF=\angle DAB=\angle CDE$. We also have $p(DCF)=p(DAE)$. Claim: $\triangle DCF\equiv\triangle DAE$. Proof: We note that $\angle FDC=\angle EDA$, the perimeters of the triangles are equal and the heights from $D$ to the (shared) bases are equal. We claim that this is enough to uniquely determine the triangle up to reflection. Suppose $\angle CDF=\theta$, and denote the height from $D$ to $AC$ to be $h$, and the side opposite $D$ to have length $a$. Using $s$ as the semiperimeter, we arrive at $$\frac{s-a}{r}=\cot\frac{\theta}{2}$$$$\frac{a}{r}=\frac{2S}{rh}=\frac{2s}{h}$$Hence, $$\frac{s}{r}=\cot\frac{\theta}{2}+\frac{2s}{h}$$This means that $r$ is uniquely determined by $s$, $\theta$ and $h$. The side length $a$ is also determined by these numbers. It is also easy to see that $a$, $h$ and $\theta$ uniquely determine a triangle. Hence, we have $\triangle DCF$ is uniquely determined by $\theta$, $h$ and $s$, which means that our lemma is proven. To finish off, $$[ADE]=[CDF]=4[ABC]$$

ACGNmath wrote: [asy][asy] size(10cm); pair A = dir(70); pair B = dir(210); pair C = dir(330); draw(A--B--C--A--cycle); pair D = 2*B-C; draw(A--D--B); pair F = 2*A-C; pair E = 2*C-A; draw(D--E--F--D--cycle); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); [/asy][/asy] It is clear that $p(ABC)<p(ADE)$, thus $2p(ABC)=p(ADE)$. Let $F$ be the reflection of $C$ across $A$. Then $\angle ADF=\angle DAB=\angle CDE$. We also have $p(DCF)=p(DAE)$. Claim: $\triangle DCF\equiv\triangle DAE$. I'll borrow the above diagram Here's another way to prove that both triangles are congruent by length bashing. Let $DE = a, DC = c, DA = d, DF = f, EC = b, AF = e$. Let $\angle EDC = \angle DAF = \theta$. Now, we have three equations: \begin{align*} a + b + c &= d + e + f \ &(\text{Perimeter}) \\ ac &= df \ &(\text{Area } = \frac{1}{2} ac \sin \theta = \frac{1}{2} df \sin \theta ) \\ \frac{a^2 + c^2 - b^2}{2ac} &= \frac{d^2 + f^2 - e^2}{2df} \ &\text{(The value of } \cos \theta) \end{align*}We now have $a^2 + c^2 - b^2 = d^2 + f^2 - e^2$. Therefore, \[ (a + b + c)(a + c - b) = (a+c)^2 - b^2 = (a^2 + c^2 - b)^2 + 2ac = (d^2 + f^2 - e^2) + 2df = (d + f)^2 - e^2 = (d + f - e)(d + e + f) \]This forces $a + c - b = d + f - e$. Hence, $b = e$ and $a + c = d + f$. However, $ac = df$. By Vieta Argument, we know that $\{ a, c \} = \{ d, f \}$. Finishing our claim. The rest is essentially the same as above's post.