∵D是BC的中点
∴BD = CD
∵AD = AD,AB = AC
∴△ABD≌△ACD
∴∠BAD = ∠CAD
∵AB = AC
∴∠B = ∠C
∵∠B + ∠C + ∠BAD + CAD = 2(∠B + ∠BAD) =180°
∴∠B+∠BAD = 90°
∴AD⊥BC
∵D是BC的中点
∴BD = CD
∵AD = AD,AB = AC
∴△ABD≌△ACD
∴∠BAD = ∠CAD
∵AB = AC
∴∠B = ∠C
∵∠B + ∠C + ∠BAD + CAD = 2(∠B + ∠BAD) =180°
∴∠B+∠BAD = 90°
∴AD⊥BC