(1)
∵∠BDP=45°=∠BAC;
∴ADPB四点共圆,∠BAD=∠DPB=90°=∠ABC;
∴AD∥BC;
(2)
由1知,PA是∠DAB角平分线,故AD:AB=1:2;令AD=a;AB=2a;BD=√5a;BP=DP=√(5/2)a;由托勒密定理,AP=3/√2a;AC=2√2a;故PC:AC=1:4;
(3)
6;由上面逆推,设AD=xAB;可得AP:AC=(1+x)/(2x),解x即可
(1)
∵∠BDP=45°=∠BAC;
∴ADPB四点共圆,∠BAD=∠DPB=90°=∠ABC;
∴AD∥BC;
(2)
由1知,PA是∠DAB角平分线,故AD:AB=1:2;令AD=a;AB=2a;BD=√5a;BP=DP=√(5/2)a;由托勒密定理,AP=3/√2a;AC=2√2a;故PC:AC=1:4;
(3)
6;由上面逆推,设AD=xAB;可得AP:AC=(1+x)/(2x),解x即可