1. **Problem statement:** Given that ABC and CDE are straight lines and AE is parallel to BD, find the sizes of angles: a) $\angle ABD$ b) $\angle AED$ c) $\angle BDC$ d) $\angle ACE$ 2. **Known angles:** - $\angle B = 36^\circ$ - $\angle A = 71^\circ$ - $\angle D = 81^\circ$ (inside triangle CDE) - $\angle A = 68^\circ$ (inside triangle CBA) - $\angle V = 28^\circ$ --- ### Step 1: Use parallel lines property Since AE is parallel to BD, alternate interior angles are equal. ### Step 2: Find $\angle ABD$ $\angle ABD$ corresponds to $\angle AED$ because AE $\parallel$ BD. ### Step 3: Find $\angle ABD$ In triangle ABD, sum of angles is $180^\circ$: $$\angle A + \angle B + \angle ABD = 180^\circ$$ $$71 + 36 + \angle ABD = 180$$ $$\angle ABD = 180 - 71 - 36 = 73^\circ$$ ### Step 4: Find $\angle AED$ By parallel lines, $\angle AED = \angle ABD = 73^\circ$ ### Step 5: Find $\angle BDC$ On straight line BDC, sum of angles is $180^\circ$: $$\angle BDC + \angle D = 180^\circ$$ $$\angle BDC = 180 - 81 = 99^\circ$$ ### Step 6: Find $\angle ACE$ On straight line ACE, sum of angles is $180^\circ$. Given $\angle A = 68^\circ$ inside triangle CBA, and $\angle ACE$ is supplementary to $\angle A$: $$\angle ACE = 180 - 68 = 112^\circ$$ --- ### Final answers: - a) $\angle ABD = 73^\circ$ - b) $\angle AED = 73^\circ$ - c) $\angle BDC = 99^\circ$ - d) $\angle ACE = 112^\circ$