F.3 equaliteral

Consider ∆PQD and ∆PCBBecause QD // CB, ∆PQD and ∆PCB are similar (all corresponding angles are equal AAA) Therefore PQ / PC = PD / PB …(1) Consider ∆PDR and ∆PBTBecause DR // BT, ∆PDR and ∆PBT are similar (all corresponding angles are equal AAA)Therefore PR / PT = PD / PB…(2)(1) & (2) => PQ / PC = PR / PTConsider ∆PQR and ∆PCTPQ / PC = PR / PT∠QPR = ∠CPT (common)Hence ∆PQR and ∆PCT are similar (SAS)=>∠PQR = ∠PCT => QR // CT (corresponding angle of // lines)