8.4. The shorter leg $AC$ of the right triangle $ABC$ has a length of $b$. A point $D$ is chosen on the hypotenuse $AB$ such that $BD = BC$. A point $E$ is taken on the leg $BC$ such that $DE = BE = m$. Find the perimeter of the quadrilateral $ADEC$.

---

Translation:
8.4. The shorter leg $AC$ of the right triangle $ABC$ has a length of $b$. A point $D$ is chosen on the hypotenuse $AB$ such that $BD = BC$. A point $E$ is taken on the leg $BC$ such that $DE = BE = m$. Find the perimeter of the quadrilateral $ADEC$.
Answer: $2 m+b$.

Solution. Let $M$ be the midpoint of the hypotenuse $AB$ of the given right triangle $ABC$. Then, $AM=BM=CM$ and $\angle BDE=\angle MBC=\angle MCB$ (angles at the base of two isosceles triangles). Since $BD=BC$, $\triangle BDE=\triangle BCM$ (by side and two adjacent angles), hence $CM=DE=m$ and $CE=BC-BE=BD-BM=DM$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_eac97ed2a88fb4e49f9dg-2.jpg?height=274&width=506&top_left_y=620&top_left_x=1409)
$P_{ADE C}=AD+DE+EC+CA=AD+m+DM+b=AM+m+b=2m+b$.