令∠BAD=α，∠CAD=β，则∠BAC=α+β
tanα=

BD

AD

=

1

3

tanβ=

CD

AD

=

3

6

=

1

2

∴tan（α+β）=

tanα+tanβ

1−tanαtanβ

=

1 2 + 1 3

1− 1 2 × 1 3

=1
又0＜∠BAC＜π，则0＜α+β＜π
所以α+β=

π

4

即∠BAC=

π

4

故答案为：

π

4