amparvardi wrote: $K, L, M$ and $N$ are points on sides $AB, BC, CD$ and $DA$, respectively, of the unit square $ABCD$ such that $KM$ is parallel to $BC$ and $LN$ is parallel to $AB$. The perimeter of triangle $KBL$ is equal to $1$. What is the area of triangle $MND$ ? Proof. Denote the midpoints $X$ , $Y$ of the sides $[AB]$ , $[BC]$ respectively. Since $BX=BY=\frac 12$ - semiperimeter of $\triangle KBL$ obtain that the incircle $w$ of the square is the $B$-exincircle of $\triangle KBL$ , i.e. $KL$ is tangent to $w$ . Denote $T\in KL\cap w$ and $KT=KX=x$ , $LT=LY=y$ . Thus, $BK=\frac 12-x$ , $BL=\frac 12-y$ and $KL=x+y$ . Since $KL^2=BK^2+BL^2\iff$ $\frac 12-(x+y)=2xy\iff$ $4xy+2(x+y)+1=2\iff$ $\boxed{(1+2x)(1+2y)=2}\ (*)$ . On other hand, $AK=DM=\frac 12+x$ , $CL=DN=\frac 12+y$ and $[MDN]=\frac 12\cdot\left(\frac 12+x\right)\left(\frac 12+y\right)$ , i.e. $[MDN]=\frac 18\cdot (1+2x)(1+2y)$ . In conclusion, using the relation $(*)$ obtain that $[MDN]=\frac 14$ .

Let the length of side $KB$ be $x$ and the length of side $BL$ be $y$. Since we're given that $KM$ is parallel to $BC$ and $NL$ is parallel to $AB$, we can conclude that side $DM=1-x$ and side $ND=1-y$. We are given that the perimeter of triangle $KBL$ is 1, thus $x+y+ \sqrt (x^2+y^2) = 1$. By rewriting this equation, we see that $2xy+1-2x-2y=0$. Now, we focus on the area of triangle $MND$. Since the length of the legs are $1-x$ and $1-y$, the area is $(1-x)(1-y)/2$ which is equal to $(1+xy-x-y)/2$. Aha, this looks very similar to the equation $2xy+1-2x-2y=0$, so we subtract $1$ on both sides, divide by $2$, and add $1$. This gives us $1+xy-x-y=1/2$. Finally, we divide by $2$ and get $\boxed {1/4}$.

See here PP7 and an easy its extension.