Question

如图,梯形ABCD中,AD//BC,∠ABC=90°,对角线AC⊥BD与P点,已知AD/BC=3/4,则BD:AC的值为？

Answer 1

解：AD//BC,∠ABC=90°,则:∠BAD=90°;且PD/PB=AD/BC=3/4.
设PD=3m,则PB=4m.
又BD⊥AC,则:⊿APB∽⊿DPA,得AP/DP=PB/AP,AP^2=DP*PB=12m^2.
AB=√(AP^2+PB^2)=√(12m^2+16m^2)=2√7m;
AD=√(AP^2+PD^2)=√(12m^2+9m^2)=√21m.
∵∠ABC=∠DAB=90°;
∠DAP=∠ABD(均为角BAP的余角).
∴⊿ABC∽⊿DAB,得:BD/AC=AD/AB=(√21m)/(2√7m)=(√3)/2.

追问

解：AD//BC,∠ABC=90°,则:∠BAD=90°;且PD/PB=AD/BC=3/4.
设PD=3m,则PB=4m.
又BD⊥AC,则:⊿APB∽⊿DPA,得AP/DP=PB/AP,AP^2=DP*PB=12m.
AB=√(AP^2+PB^2)=√(12m+16m)=2√7m;
AD=√(AP^2+PD^2)=√(12m+9m)=√21m.
∵∠ABC=∠DAB=90°;
∠DAP=∠ABD(均为角BAP的余角).
∴⊿ABC∽⊿DAB,得:BD/AC=AD/AB=(√21m)/(2√7m)=(√3)/2.

把答案变成以上，就给分