Question

consider the following figure.
(note that the figure is not drawn to scale.)
order the side lengths ( bc ), ( ce ), ( bd ), ( cd ), and ( ed ) from least to greatest.
\\( \square < \square < \square < \square < \square \\)

Explanation:

Step1: Analyze triangle BCD

In $\triangle BCD$, angles are $\angle B = 46^\circ$, $\angle CDB = 69^\circ$. So $\angle BCD = 180 - 46 - 69 = 65^\circ$. By the Law of Sines, side lengths are proportional to the sines of opposite angles. So $BC$ (opposite $69^\circ$), $BD$ (opposite $65^\circ$), $CD$ (opposite $46^\circ$). So $CD < BD < BC$.

Step2: Analyze triangle CDE

In $\triangle CDE$, angles are $\angle E = 64^\circ$, $\angle DCE = 55^\circ$. So $\angle CDE = 180 - 64 - 55 = 61^\circ$. By Law of Sines, $CE$ (opposite $61^\circ$), $CD$ (opposite $64^\circ$), $ED$ (opposite $55^\circ$). So $ED < CE < CD$.

Step3: Combine both triangles

From $\triangle CDE$: $ED < CE < CD$. From $\triangle BCD$: $CD < BD < BC$. So combining, $ED < CE < CD < BD < BC$.

Answer:

$ED < CE < CD < BD < BC$