设AC和BD交于O
∵ABCD是菱形
∴AC⊥BD
AO=1/2AC=1/2×10=5
BO=1/2BD=1/2×6=3
∠BAC=∠CAD=1/2∠BAD
∴△ABO是直角三角形
∴tan∠BAO
=tan∠BAC
=tan(∠BAD/2)
=BO/AO
=3/5
即tan(∠BAD/2)=3/5
设AC和BD交于O
∵ABCD是菱形
∴AC⊥BD
AO=1/2AC=1/2×10=5
BO=1/2BD=1/2×6=3
∠BAC=∠CAD=1/2∠BAD
∴△ABO是直角三角形
∴tan∠BAO
=tan∠BAC
=tan(∠BAD/2)
=BO/AO
=3/5
即tan(∠BAD/2)=3/5