Question

(i): Given triangles ABC and PQR with $\angle A=\angle P$, $\angle C=\angle R$, and AD perpendicular to BC, PS perpendicular to QR. Since two angles are equal, the third angles are also equal: $\angle B=\angle Q$. Therefore, $△ABC\sim △PQR$ by AA similarity. Since AD and PS are corresponding altitudes from equal angles, $\angle ADB=\angle PSQ=90°$. Thus, $△ADB\sim △PSQ$ by AA similarity. | (ii): Since $△ADB\sim △PSQ$, the corresponding sides are proportional: $\frac{AD}{PS}=\frac{BD}{QS}$. Cross-multiplying gives $AD\times QS=BD\times PS$.