1. **Problem statement:** We have two intersecting lines AB and CD at point E. Given: $|AE|=25$, $|CE|=24$, $|EB|=30$, and $\angle ACE=90^\circ$. Find: (i) $|\angle AEC|$ to 2 decimal places. (ii) Area of triangle ACE. (iii) Given area of triangle EBD equals area of ACE, find $|ED|$ to nearest whole number. 2. **Step (i): Find $|\angle AEC|$** Triangle ACE is right angled at C. Calculate $AC = \sqrt{25^2 - 24^2} = 7$. Use cosine rule: $$625 = 49 + 576 - 2 \times 7 \times 24 \times \cos(\angle AEC)$$ Simplify: $$625 = 625 - 336 \cos(\angle AEC)$$ $$0 = -336 \cos(\angle AEC)$$ Divide both sides by $\cancel{336}$: $$0 = -\cancel{336} \cos(\angle AEC) / \cancel{336}$$ $$\cos(\angle AEC) = 0$$ So, $$|\angle AEC| = 90^\circ$$ 3. **Step (ii): Area of triangle ACE** Since $\angle ACE=90^\circ$, area is: $$\frac{1}{2} \times 25 \times 24 = 300$$ 4. **Step (iii): Find $|ED|$ given area of triangle EBD equals area of ACE** Area EBD = 300, base $EB=30$. $$300 = \frac{1}{2} \times 30 \times ED$$ $$300 = 15 \times ED$$ Divide both sides by $\cancel{15}$: $$\frac{300}{\cancel{15}} = \cancel{15} ED / \cancel{15}$$ $$20 = ED$$ **Final answers:** (i) $|\angle AEC| = 90^\circ$ (ii) Area of triangle ACE = 300 (iii) $|ED| = 20$