(2)∵AP=
AD,△ABP和△ABD的高相等,
∴S △ABP=
,
又∵PD=AD-AP=
AD,△CDP和△CDA的高相等,
∴S △CDP=
,
∴ S △PBC=S 四边形ABCD-S △ABP-S △CDP
=S 四边形ABCD-
S △ABD-
S △CDA
=S 四边形ABCD-
(S 四边形ABCD-S △DBC)-
(S 四边形ABCD-S △ABC)
=
S △DBC+
S △ABC
∴S △PBC=
S △DBC+
S △ABC;
(3)S △PBC=
S △DBC+
S △ABC;
(4)S △PBC=
S △DBC+
S △ABC;
∵AP=
AD,△ABP和△ABD的高相等,
∴S △ABP=
S △ABD
又∵PD=AD-AP=
AD,△CDP和△CDA的高相等,
∴S △CDP=
S △CDA,
∴S △PBC=S 四边形ABCD-S △ABP-S △CDP
=S 四边形ABCD-
S △ABD-
S △CDA
=S 四边形ABCD-
(S 四边形ABCD-S △DBC)-
(S 四边形ABCD-S △ABC)
=
S △DBC+
S △ABC,
∴S △PBC=