Cho hình chữ nhật ABCD có cạnh AD = a, M là trung điểm

Ta có: cos\(\widehat {MDB} = \sqrt {1 - {{\left( {\frac{1}{3}} \right)}^2}} = \frac{{2\sqrt 2 }}{3}\)

Xét tam giác AMD có: S_AMD = \(\frac{1}{2}\)AD.AM

S_AMD = \(\frac{1}{2}.AD.\frac{{AB}}{2} = \frac{1}{4}.AD.AB = \frac{1}{4}.AD.a\)

Xét tam giác MDB có: S_BMD = \(\frac{1}{2}\)MD.DB.sin\(\widehat {MDB} = \frac{1}{6}.MD.DB\)

Xét tam giác ABD có: S_ABD = \(\frac{1}{2}\)AD.AB = \(\frac{1}{2}.a.AB\) (1)

Mà S_AMD + S_BMD = S_ABD

⇔ \(\frac{1}{4}.a.AB + \frac{1}{6}.MD.DB = \frac{1}{2}.a.AB\)

Xét tam giác MDB có: MB² = MD² + DB² – 2.MB.BD.cos\(\widehat {MDB}\)

⇔ \(\frac{{A{B^2}}}{4} = M{D^2} + D{B^2} - 2.MB.BD.\frac{{2\sqrt 2 }}{3}\)

⇔ MD.DB = \(\left( {M{D^2} + D{B^2} - \frac{{A{B^2}}}{4}} \right).\frac{3}{{4\sqrt 2 }}\)

⇔ MD.DB = \(\left( {2{a^2} + A{B^2}} \right).\frac{3}{{4\sqrt 2 }}\left( 2 \right)\)

Thế (2) vào (1) ta có:

\(\frac{1}{4}a.AB + \frac{{\sqrt 2 }}{{16}}\left( {2{a^2} + A{B^2}} \right) = \frac{1}{2}.AB.a\)

⇔ \(\frac{{\sqrt 2 }}{{16}}\left( {2{a^2} + A{B^2}} \right) = \frac{1}{4}a.AB\)

⇔ AB² – 2\(\sqrt 2 \)a.AB + 2a² = 0

⇔ (AB – a\(\sqrt 2 \))² = 0

⇔ AB = a\(\sqrt 2 \).