问题 填空题
设D、P为△ABC内的两点,且满足
AD
=
1
5
(
AB
+
AC
)
AP
=
AD
+
1
10
BC
,则
S△APD
S△ABC
=______.
答案

取BC的中点E,连接AE,则

AB
 +
AC
=2
AE
AD
=
2
5
AE
;∵
AP
=
AD
+
1
10
BC
AP
-
AD
=
1
10
BC
DP
=
1
10
BC
故DPBC且DP=
1
10
BC∴△APD与△ABC的高之比为h:H=AD:AE=2:5

S△APD:S△ABC=

h
H
×
DP
BC
=
2
5
×
1
10
=
1
25

故答案为:

1
25

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