D为三角形ABC的边BC上一点,且角BAD=角C,试证明：AD的平方/AC的平方=BD/BC?

∵∠ABD=∠C,∠B=∠B∴△ABD∽△CBA∴AD/AC=AB/BC=BD/AB∴AB^2=BD *BC∴AD^2/AC^2=AB^2/BC^2=BD*BC/BC^2=BD/BC 面积法∵∠ABD=∠C,∠B=∠B∴△ABD∽△CBA∴S△ABD：△SABC=BD∶CD=AD^2∶AC^2 (面积比等于底的比,面积...