4. Let $ABCD$ be a parallelogram where $AB > BC$, $[AE$ is the bisector of angle $A$, $[BF$ is the bisector of angle $B$ ($E, F \in (DC)$), $X$ is the midpoint of segment $[AE]$, and $Y$ is the midpoint of segment $[BF]$.

a) Prove that $DXYF$ is a parallelogram.

b) If $5AD = 3AB$ and $XY = 24 \text{ cm}$, find the perimeter of parallelogram $ABCD$.
Solution:
a) $\triangle \mathrm{ADE}$ and $\triangle \mathrm{BCF}$ are isosceles $\rightarrow \mathrm{DE}=\mathrm{AD}=\mathrm{BC}=\mathrm{CF} \rightarrow \mathrm{DF}=E C$.

$\mathrm{ABEF}$ is a trapezoid and $\mathrm{X}, \mathrm{Y}$ are midpoints $\rightarrow \mathrm{XY}\|\mathrm{AB}\| \mathrm{DF} \rightarrow \mathrm{XY}=\frac{A B-E F}{2}=\frac{D C-E F}{2}=D F$.

$D E=A D=C F=A B \rightarrow D F=E C \rightarrow D X Y F$ is a parallelogram.

(4 points)

b) Let $E F=x \rightarrow D F=E C=24 \rightarrow A B=48+x, A D=24+x$

$5(24+x)=3(48+x) \rightarrow x=12 \rightarrow P_{A B C D}=192 \text{ cm}$.

(3 points)