如图,已知D是△ABC的边BC上一点,且AC^2-DC^2=AD^2,求证:AB^2-AC^2=BD^2-cd^2

问题描述:

如图,已知D是△ABC的边BC上一点,且AC^2-DC^2=AD^2,求证:AB^2-AC^2=BD^2-cd^2

勾股定理嘛,由AC^2-DC^2=AD^2可知,AD⊥BC,所以AB^2-BD^2=AD^2,所以AC^2-DC^2=AB^2-BD^2,移项可得结果