Fresh day Kwanga, let's solve. To prove that $\angle DAC = \angle ABC$, we need to show that $\triangle ABC$ is similar to $\triangle DAC$. We will use the Side-Angle-Side (SAS) similarity criterion. This requires two pairs of corresponding sides to be proportional, and the included angles between those sides to be equal. Let's list the side lengths for the two triangles: For $\triangle ABC$: $AB = 10 \, \text{m}$ $AC = 7 \, \text{m}$ $BC = 10.2 \, \text{m}$ For $\triangle DAC$: $DA = 14 \, \text{m}$ $AC = 7 \,