a) Đặt AC = b ; AB = c
Ta có bc = 2SABC = 2SADB + 2SADC
= AB.ADsin45o + AC.ADsin45o
= (b + c)ADsin45o
bc = (b + c)AD.\(\dfrac{\sqrt{2}}{2}\)
=> \(\dfrac{AD}{\sqrt{2}}=\dfrac{bc}{b+c}=>\dfrac{\sqrt{2}}{AD}=\dfrac{b=c}{bc}=\dfrac{1}{c}+\dfrac{1}{b}\)
Hay \(\dfrac{\sqrt{2}}{AD}=\dfrac{1}{AB}+\dfrac{1}{AC}\)

