$\triangle ABC$. Let $X\in BC; Y\in CA, Z\in AB$ such that $\angle YXZ= \angle BAC, \angle XZY=\angle ACB, \angle ZYX=\angle CBA$.
Prove that: $S_{XYZ}\geq \frac{1}{4}S_{ABC}$
P/s: I have proved that the length of circumscribed circles of $\triangle AYZ,\triangle BXZ,\triangle CXY,\triangle XYZ$ are the same
For simplicity of description, let us choose the coordinate system so that $A$ is the origin, $AB$ is in the direction of $x$-axis and $C$ is in the upper half-plane. Let
The restriction on angles
$$
\begin{cases}
\angle YXZ &= \angle BAC = \alpha\\
\angle ZYX &= \angle CBA = \beta\\
\angle XZY &= \angle ACB = \gamma
\end{cases}$$
tell us $\triangle XYZ$ is similar to $\triangle ABC$. We can obtain the former by scaling $\triangle ABC$ with a scalar factor $\rho$, rotate it counter-clockwisely for an angle $\pi + \phi$ and then translate it to its final places.
The resulting configuration is illustrated in following diagram:
As seen from the digram, the length of the sides $XY$, $YZ$, $ZX$ is $\rho a$, $\rho b$, $\rho c$. As a result, we have
$$\verb/Area/(\triangle XYZ) = \rho^2\verb/Area/(\triangle ABC)$$
Our task reduced to showing $\rho \ge \frac12$.
Let $\mu = \frac{|AZ|}{|AB|}$. Applying the sine rule to triangles $\triangle AZY$ and $\triangle ZBX$, we have
$$\frac{\rho a}{\sin\alpha} = \frac{|YZ|}{\sin\angle YAZ} = \frac{|AZ|}{\sin\angle ZYA} = \frac{\mu c}{\sin(\gamma+\phi)}\\ \frac{\rho b}{\sin\beta} = \frac{|XZ|}{\sin\angle ZBX} = \frac{|ZB|}{\sin\angle BXZ} = \frac{(1-\mu)c}{\sin(\gamma-\phi)} $$ Together with the sine rule from $\triangle ABC$,
$$\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma}$$ We find $$ \begin{cases} \mu &= \rho\frac{\sin(\gamma + \phi)}{\sin\gamma},\\ 1-\mu &= \rho\frac{\sin(\gamma-\phi)}{\sin\gamma} \end{cases} \quad\implies\quad 1 = \frac{\rho}{\sin\gamma}(\sin(\gamma + \phi) + \sin(\gamma - \phi)) = 2\rho\cos\phi$$
This leads to $\displaystyle\;\rho = \frac{1}{2\cos\phi} \ge \frac12$ and we are done!
We may consider the following configuration:
Given $XYZ$, we take $X'Y'Z'$ as the anticomplementary triangle of $XYZ$. Then we take $\Gamma_A,\Gamma_B,\Gamma_C$ as the circumcircles of $X'YZ,XY'Z,XYZ'$ and $A\in\Gamma_A$. We take $B$ as $AZ\cap\Gamma_B$ and $C$ as $BX\cap\Gamma_C$. $A,X,C$ will be on the same line, and $\Gamma_A,\Gamma_B,\Gamma_C$ will concur in the orthocenter $H$ of $XYZ$.
The ratio $\frac{[ABC]}{[XYZ]}$ will depend just on the position of $A$ on $\Gamma_A$. If $x=YZ,y=XZ,z=XY$ and $R_A,R_B,R_C$ are the circumradii of $\Gamma_A,\Gamma_B,\Gamma_C$, then $R_A=\frac{x}{2\sin A}$ and so on, so $\Gamma_A,\Gamma_B,\Gamma_C$ and the circumcircle $\Gamma$ of $XYZ$ are congruent. That gives that all the green segments in the following picture have the same length:
Given the last decomposition, isn't it trivial that $\frac{[ABC]}{[XYZ]}\leq 4$? Just write the area of every small triangle with two green sides as $\frac{R^2}{2}$ times the sine of something. Equality occurs just when $AX,BY,CZ$ concur, i.e. just when $XYZ$ is the medial triangle of $ABC$. The claim just follows from:
$$ 2\sin(2A)-\sin(2A+2\theta)-\sin(2A-2\theta) = 4\sin^2(\theta)\sin(2A)\geq 0.$$
