In the opposite figure \( \overrightarrow{\mathrm{DC}} \) bisects \( \angle \mathrm{ACB} \) , \( \mathrm{AE}=12 \mathrm{~cm} \). , \( \mathrm{CE}=6 \mathrm{~cm} \). \( \mathrm{BC}=9 \mathrm{~cm} \). rove that \( : \overline{\mathrm{DE}} / / \overline{\mathrm{BC}} \)

To prove that \( \overline{\mathrm{DE}} \parallel \overline{\mathrm{BC}} \), we can use the Angle Bisector Theorem. This theorem states that if a ray bisects an angle of a triangle, it divides the opposite side into two segments that are proportional to the other two sides. In this case, \( \overrightarrow{\mathrm{DC}} \) bisects \( \angle \mathrm{ACB} \), so we have: \[ \frac{\mathrm{AE}}{\mathrm{CE}} = \frac{\mathrm{AB}}{\mathrm{BC}} \] Substituting the known values: \[ \frac{12}{6} = \frac{\mathrm{AB}}{9} \] This simplifies to: \[ 2 = \frac{\mathrm{AB}}{9} \] Thus, \( \mathrm{AB} = 18 \mathrm{cm} \). According to the properties of ratios and similar triangles, since \( \overline{\mathrm{DE}} \) and \( \overline{\mathrm{BC}} \) are created by a transversal intersecting two parallel lines, we can conclude that \( \overline{\mathrm{DE}} \parallel \overline{\mathrm{BC}} \). Hence proved!