標題:
F.3 equaliteral
發問:
圖片參考：http://imgcld.yimg.com/8/n/HA00654156/o/701103120125613873389240.jpg Prove that QR//CT Please show clear step:)
最佳解答:
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)
F.3 equaliteral
發問:
圖片參考：http://imgcld.yimg.com/8/n/HA00654156/o/701103120125613873389240.jpg Prove that QR//CT Please show clear step:)
最佳解答:
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)
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