1. **State the problem:** We have a right-angled triangle ABC with right angle at C. Lines AEC and ADB are straight, and ED is parallel to CB. We need to prove triangle ABC is similar to triangle ADE (part a), and then find the length of EC (part b). 2. **Prove similarity of triangles ABC and ADE:** - Since ED is parallel to CB, corresponding angles are equal: \(\angle ADE = \angle ABC\). - Both triangles share \(\angle A\). - Therefore, by AA (Angle-Angle) similarity criterion, \(\triangle ABC \sim \triangle ADE\). 3. **Use similarity to find EC:** - From similarity, corresponding sides are proportional: $$\frac{AB}{AD} = \frac{BC}{DE} = \frac{AC}{AE}$$ - Given: \(AC = 18\) cm, \(BC = 30\) cm, \(CB = 30\) cm, \(ED = 20\) cm. - Using the ratio of sides corresponding to BC and ED: $$\frac{BC}{ED} = \frac{30}{20} = \frac{3}{2}$$ - So, the scale factor from \(\triangle ADE\) to \(\triangle ABC\) is \(\frac{3}{2}\). 4. **Find AE:** - Since \(\frac{AC}{AE} = \frac{3}{2}\), then $$AE = \frac{2}{3} \times AC = \frac{2}{3} \times 18 = 12 \text{ cm}$$ 5. **Find EC:** - Since E lies on AC, and AC = 18 cm, $$EC = AC - AE = 18 - 12 = 6 \text{ cm}$$ **Final answer:** $$EC = 6 \text{ cm}$$