Cho tam giác ABC vuông tại \(A.\) Chứng minh rằng \[\tan \frac{{\widehat {ABC}}}{2} = \frac{{AC}}{{AB + BC}}\]

a) Xét \[\Delta AMB\] vuông tại \[B\]có: O10-2024-GV154.\[AM = \sqrt {B{A^2} - B{M^2}}  = \sqrt {{5^2} + {{12}^2}}  = 13cm\]

Ta có: O10-2024-GV154 \[\sin \widehat {AMB} = \frac{{AB}}{{AM}} = \frac{{12}}{{13}}\];\[co{\mathop{\rm s}\nolimits} \widehat {AMB} = \frac{{BM}}{{AM}} = \frac{5}{{13}}\];\[\tan \widehat {AMB} = \frac{{AB}}{{BM}} = \frac{{12}}{5}\];\[\cot \widehat {AMB} = \frac{1}{{{\rm{tan}}\widehat {AMB}}} = \frac{5}{{12}}\].

b)Xét \[\Delta AMB = \Delta DNA(c.g.c)\]\[ \Rightarrow AM = DN\].

c)Ta có: O10-2024-GV154 \[\tan \widehat {BAM} = \frac{{BM}}{{AB}}\];\[\cot \widehat {AND} = \frac{{AN}}{{AD}}\].Do\[ \Rightarrow BM = AN;AB = A{\rm{D}}\]

\[ \Rightarrow \tan \widehat {BAM} = \cot \widehat {AND}\] nên trong \[\Delta AKN\] có: O10-2024-GV154.\[\widehat {BAM}\] và\[\widehat {AND}\]phụ nhau.Suy ra \[\widehat {AKN} = {90^0}\].Vậy\[{\rm{AM}} \bot {\rm{DN}}\].

Ta có \(\cos A = \frac{{AE}}{{AB}};\,\,\cos B.cosC = \frac{{BD}}{{AB}}.\frac{{DC}}{{AC}}\,\,\,\,\,\,\left( 1 \right)\)

Mặt khác

\( \Rightarrow \frac{{DB}}{{DA}} = \frac{{DH}}{{DC}} \Rightarrow DB.DC = DA.DH\)

Hay \(DB.DC = DA.AH\,\,\left( 2 \right)\)

Mà \( \Rightarrow \frac{{AE}}{{AD}} = \frac{{AH}}{{AC}} \Rightarrow AE.AC = AD.DH\,\,\left( 3 \right)\)

Từ \(\left( 1 \right),\,\left( 2 \right),\,\left( 3 \right)\) ta có \({\mathop{\rm cosB}\nolimits} .cosC = \frac{{AE.AC}}{{AB.AC}} = \frac{{AE}}{{AB}}\) suy ra \[\cos A = \cos B.cosC\]