Problem - 3042
Three squares are drawn on the sides of $\triangle{ABC}$ (i.e. the square on $AB$ has $AB$ as one of its sides and lies outside $\triangle{ABC}$). Show that the lines drawn from the vertices $A, B, C$ to the centers of the opposite squares are concurrent.
Let $A', B', C'$ be the centers of the squares opposite $A$, $B$, $C$, respectively. Then $\triangle{ABA'}$ and $\triangle{CBC'}$ have the same area because
$$\angle{ABA'} =\angle{CBC'}=\angle{ABC}+45^\circ$$
and $$\frac{BA'}{BC} = \frac{BC'}{AB} =\frac{\sqrt{2}}{2}$$
mean $$\frac{1}{2}\cdot BA\cdot BA'\sin\angle{ABA'}=\frac{1}{2}\cdot BC\cdot BC'\sin\angle{CBC'}\implies S_{\triangle{ABA'}}=S_{\triangle{CBC'}}$$
Similarly, it must be true that
$$S_{\triangle{CAC'}}=S_{\triangle{BAB'}}\quad\text{and}\quad S_{\triangle{ACA'}}=S_{\triangle{BCB'}}$$
Let $AA'$, $BB'$,$CC'$ intersect with $BC$, $CA$, $AB$ at $X$, $Y$, $Z$, respectively. Then
\begin{align}
\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA} = \frac{S_{\triangle{CAC'}}}{S_{\triangle{CBC'}}}\cdot\frac{S_{\triangle{ABA'}}}{S_{\triangle{ACA'}}}\cdot\frac{S_{\triangle{BCB'}}}{S_{\triangle{BAB'}}}=1
\end{align}
Hence, by the Ceva's theorem, $AA'$, $BB'$ and $CC'$ are concurrent.