Question

已知角AOB=90度,OM是角AOB的平分线,点P是OM上的任意一点

已知∠AOB=90°,OM是∠AOB的平分线,点P是OM上的任意一点,点D是OB上的点连接PD,过点P做PC⊥PD,交直线OA于点C,连接CD交OM于点G (1)求证PC=PD (2)若PG=（根号3）PD/2,求△PDO与△PDG的面积比

Answer 1

.....................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................................................................................................................................................................................................................................................................

Answer 2

解：
1）作PE⊥OA于E，PF⊥OB于F，
∵OM是∠AOB的平分线
∴PE=PF
∵∠AOB=90°
∴PEOF是正方形
∵PC⊥PD
∴∠EPC+∠CPF=∠CPF+∠FPD
∴∠EPC=∠FPD
∴Rt△PEC≌Rt△PFD（HL）
∴PC=PD
2）
∵PC⊥PD，PC=PD
∴∠PDG= 45°
∵∠AOB=90°，OM是∠AOB的平分线
∴∠POD= 45°
∵∠DPG=∠OPD
∴△DPG∽△OPD
∴S△POD/S△PDG=（PD/PG）^2
=［PD/（√3PD/2）］^2
=1/(3/4)
=4/3

Answer 3

解：
1）作PE⊥OA于E，PF⊥OB于F，
∵OM是∠AOB的平分线
∴PE=PF
∵∠AOB=90°
∴PEOF是正方形
∵PC⊥PD
∴∠EPC+∠CPF=∠CPF+∠FPD
∴∠EPC=∠FPD
∴Rt△PEC≌Rt△PFD（HL）
∴PC=PD
2）
∵PC⊥PD，PC=PD
∴∠PDG= 45°
∵∠AOB=90°，OM是∠AOB的平分线
∴∠POD= 45°
∵∠DPG=∠OPD
∴△DPG∽△OPD
∴S△POD/S△PDG=（PD/PG）^2
=［PD/（√3PD/2）］^2
=1/(3/4)
=4/3