等腰△ABC中,AB=AC,AH⊥BC于H,D是底边上任意一点,过D作BC的垂线交AC于M,交BA的延长线于N,求证：DM+DN=2AH

Question

等腰△ABC中,AB=AC,AH⊥BC于H,D是底边上任意一点,过D作BC的垂线交AC于M,交BA的延长线于N,求证：DM+DN=2AH

Answer 1

过A做AF⊥DN,交MN于F
∵AB = AC
∴∠B = ∠C
∵AH⊥BC,DN⊥BC
∴∠N = ∠CMD
∵∠CMD = ∠AMN
∴∠N = ∠AMN
∴AN =AM
∵AF⊥MN
∴F是MN的中点
∵AF⊥BC
∴四边形AHDF是长方形
∴AH = DM + MF = FN + DM
∴2AH =DM + MF + FN + DM = DM + DN