a) In the figure below, $A D=D C, A E=B D, \angle A E C=90^{\circ}$. Determine the value of the angle $\angle C B D$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-18.jpg?height=463&width=674&top_left_y=2047&top_left_x=588)
b) In the triangle $\triangle A B C$ below, $B D$ is the bisector of the angle $\angle A B C, E$ is the midpoint of $A C$, and $\angle A D B=90^{\circ}$. If $A B=12 \mathrm{~cm}$ and $B C=20 \mathrm{~cm}$, calculate the length of the segment $D E$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-19.jpg?height=534&width=990&top_left_y=424&top_left_x=630)
Solution

a) From point $D$, draw the segment perpendicular to side $B C$ as indicated in the figure below. Since $D$ is the midpoint of $A C$ and $D F \| A E$, we can conclude that $D F$ is the midline of triangle $\triangle A E C$ with respect to base $A E$. Therefore, if $A E=2 x$, then $D F=A E / 2=x$. Thus, $\sin \angle D B C=D F / B D=x / 2 x=1 / 2$. Consequently, $\angle D B C=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-19.jpg?height=368&width=554&top_left_y=1461&top_left_x=842)

b) Extend segment $A D$ until it intersects side $B C$ at $F$. Since $B D$ is the bisector of $\angle A B F$, it follows that triangles $\triangle A B D$ and $\triangle B D F$ have the same angles and one common side. Therefore, they are congruent and $B F=A B=12 \mathrm{~cm}$. Thus, $F C=8 \mathrm{~cm}$. Moreover, since $A D=D F$ and $A E=E C$, we can conclude that $D E$ is the midline of triangle $\triangle A F C$, that is, $D E=F C / 2=4 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-19.jpg?height=478&width=898&top_left_y=2171&top_left_x=679)