Question

在△ABC中.AB⊥AC,AD⊥BC于D,求证:1/AD^2=1/AB^2+1/AC^2

Answer 1

BCXAD=ABXAC（三角形面积相等）

1/AB^2+1/AC^2=(AB^2+AC^2)/(ABXAC)^2

AB^2+AC^2=BC^2,BCXAD=ABXAC

(AB^2+AC^2)/(ABXAC)^2=BC^2/(BCXAD)^2=1/AD^2

1/AB^2+1/AC^2=(AB^2+AC^2)/(ABXAC)^2=BC^2/(BCXAD)^2=1/AD^2