已知：如图，△ABC中，AB="AC," D、E在BC边上，且AD=AE，求证：BD=C

证明：作AF⊥BC于F, ∵ AB="AC," AF⊥BC于F,∴ BF=CF∵ AD="AE," AF⊥BC于F,∴ DF=EF, ∴ BF-DF=CF-EF即 BD=CE 先过A作AF⊥BC于F，而AB=AC，根据三角形三线合一定理可得BF=CF，同理可得DF=EF，再根据等式性质易证BD=CE．