∵∠BAC=90°,AB=3,AC=4,
∴BC=
AB 2 +AC 2 =
3 2 +4 2 =5,
∴BC边上的高=3×4÷5=
12
5 ,
∵AD平分∠BAC,
∴点D到AB、AC上的距离相等,设为h,
则S △ABC=
1
2 ×3h+
1
2 ×4h=
1
2 ×5×
12
5 ,
解得h=
12
7 ,
S △ABD=
1
2 ×3×
12
7 =
1
2 BD?
12
5 ,
解得BD=
15
7 .
故选A.
∵∠BAC=90°,AB=3,AC=4,
∴BC=
AB 2 +AC 2 =
3 2 +4 2 =5,
∴BC边上的高=3×4÷5=
12
5 ,
∵AD平分∠BAC,
∴点D到AB、AC上的距离相等,设为h,
则S △ABC=
1
2 ×3h+
1
2 ×4h=
1
2 ×5×
12
5 ,
解得h=
12
7 ,
S △ABD=
1
2 ×3×
12
7 =
1
2 BD?
12
5 ,
解得BD=
15
7 .
故选A.