Problem 2. a) Point $C$ is the midpoint of segment $(A B)$, $D$ is the midpoint of $(B C)$, $E$ is the midpoint of $(C D)$, and $F$ is a point on the ray $(E B)$ such that $(E B) \equiv (B F)$. If $E B=3 \text{ cm}$, calculate $A F$.
Solution. We can denote $ED = x$ and we successively have that $CD = BD = 2x$ and $EB = BF = 3x$. Since $EB = 3$ cm, we have that $x = 1$ cm. $1$ p

![](https://cdn.mathpix.com/cropped/2024_06_07_6b23f7230df04ff03783g-1.jpg?height=66&width=1719&top_left_y=2423&top_left_x=228)

Then $AF = 4$ cm + $2$ cm + $2$ cm + $3$ cm = $11$ cm $1$ p
b) Consider two adjacent angles $\angle AOB$ and $\angle BOC$ with measures $108^\circ$ and $68^\circ$ respectively. The rays $[OM$, $[ON$, and $[OP$ are the bisectors of angles $\angle AOB$, $\angle BOC$, and $\angle MON$ respectively. On the ray opposite to $[OP$, point $D$ is considered, and within the angle $\angle AOD$, point $E$ is chosen such that $m(\angle EOD) = 10^\circ$. Show that points $B$, $O$, and $E$ are collinear.

![](https://cdn.mathpix.com/cropped/2024_06_07_6b23f7230df04ff03783g-2.jpg?height=68&width=1714&top_left_y=523&top_left_x=228)$\qquad$ $\qquad$
But $P$, $O$, and $D$ are collinear and $\angle BOP \equiv EOD \Rightarrow B$, $O$, and $E$ are collinear. .................................................... $1$ p