Cho tam giác ABC vuông tại A, đường cao AH.

Chứng minh rằng: S_BHD = \(\frac{1}{4}\)S_BKC cos²ABD.

Ta có: S_ABC  = \(\frac{1}{2}CH.AB = \frac{1}{2}AC.AB.\frac{{CH}}{{AC}} = \frac{1}{2}AC.AB.\sin A\)

S_BHD = \(\frac{1}{2}BH.BD.\sin \widehat {DBH}\)

S_BKC = \(\frac{1}{2}BK.BC.\sin \widehat {KBC}\)

\(\frac{{{S_{BHD}}}}{{{S_{KBC}}}} = \frac{{BH.BD}}{{BK.BC}} = \frac{2}{8}\,.\,\frac{{BD}}{{BK}} = \,\frac{1}{4}\,.\,\frac{{B{D^2}}}{{BK.BD}}\, = \,\frac{1}{4}\,.\,\frac{{B{D^2}}}{{B{A^2}}} = \frac{1}{4}\,.\,{\cos ^2}\widehat {ABD}\).