Bài 102454

Question

CMR nếu

$ABC$ thỏa mãn

$1$ trong các điều kiện sau đây thì nó là tam giác cân

$1/$

$(a - b + c)\cot \frac{C}{2} = (a + b + c)tan\frac{B}{2}$

$2/$

$a(\cot \frac{C}{2} - tanA) = b(tanB - \cot \frac{C}{2})$

$3/$

$a + b = tan\frac{C}{2}(atanA + btanB)$

$4/$

$tanA + tanB = 2\cot \frac{C}{2}$

$5/$

$\frac{1}{2}(tanA + tanB) = \frac{{\sin A + \sin B}}{{\cos A + \cos B}}$

$6/$

$\sin A\sin B = cos^2\frac{C}{2}$

$7/$

$\frac{{\sin A + \sin B + \sin C}}{{\sin A + \sin B - \sin C}} = \cot \frac{A}{2}\cot \frac{C}{2}$

$8/$

$\sin \frac{A}{2}cos^3\frac{B}{2} = \sin \frac{B}{2}cos^3\frac{A}{2}$

$9/$

$\sin \frac{A}{2} = \frac{a}{{2\sqrt {bc} }}$

$10/$

$tan\frac{A}{2} = \frac{a}{{\sqrt {(a + b + c)(b + c - a)} }}$

$11/$

$l_a^2 = \frac{{4bc - {a^2}}}{4}$

$12/$

${a^2} + {b^2} = tan\frac{C}{2}({a^2}\tan A + b^2\tan B)$

$13/$

$h_a = \sqrt {p(p - a)}$

score 0 · Answer 1

$1/$ ta có

$(a - b + c)\cot g\frac{C}{2} = (a + b + c)tg\frac{B}{2}$

$\Leftrightarrow (\sin A + \sin C - \sin B)\cot g\frac{C}{2} = (\sin A + \sin B + \sin C)tg\frac{B}{2} (1)$
Áp dụng công thức qua các bước sau :
Trong tam giác

$ABC$ ,ta có :

$\left\{ \begin{array}{l} \sin A + \sin B + \sin C = 4\cos \frac{A}{2}c{\rm{os}}\frac{B}{2}c{\rm{os}}\frac{C}{2}\\ \sin A + \sin C - \sin B = 4\cos \frac{B}{2}\sin \frac{A}{2}\sin \frac{C}{2} \end{array} \right.$
Vì thế,(

$1$ ) tương đương

$c{\rm{os}}\frac{B}{2}\sin \frac{A}{2}\sin \frac{C}{2}\frac{{c{\rm{os}}\frac{C}{2}}}{{\sin \frac{C}{2}}} = c{\rm{os}}\frac{A}{2}c{\rm{os}}\frac{B}{2}c{\rm{os}}\frac{C}{2}\frac{{\sin \frac{B}{2}}}{{c{\rm{os}}\frac{B}{2}}}$

$\begin{array}{l} \Leftrightarrow c{\rm{os}}\frac{B}{2}\sin \frac{A}{2}c{\rm{os}}\frac{C}{2} = c{\rm{os}}\frac{A}{2}c{\rm{os}}\frac{C}{2}\sin \frac{B}{2}\\ \Leftrightarrow \sin \frac{A}{2}c{\rm{os}}\frac{B}{2} = \sin \frac{B}{2}c{\rm{os}}\frac{A}{2}(c{\rm{os}}\frac{C}{2} > 0) \end{array}$

$\begin{array}{l} \Leftrightarrow \sin \frac{{A - B}}{2} = 0\\ \Leftrightarrow A = B \end{array}$
Ta có

$DPCM$

$2/$ ta có :

$a(\cot g\frac{C}{2} - tgA) = b(tgB - \cot g\frac{C}{2})$

$\begin{array}{l} \Leftrightarrow a(tg\frac{{A + B}}{2} - tgA) = b(tgB - tg\frac{{A + B}}{2})\\ \Leftrightarrow \sin A\frac{{\sin (\frac{{A + B}}{2} - A)}}{{c{\rm{os}}\frac{{A + B}}{2}\cos A}} = \sin B\frac{{\sin (B - \frac{{A + B}}{2})}}{{\cos Bc{\rm{os}}\frac{{A + B}}{2}}}\\ \Leftrightarrow \sin A\frac{{\sin \frac{{B - A}}{2}}}{{\cos A}} = \sin B\frac{{\sin \frac{{B - A}}{2}}}{{\cos B}}\\ \Leftrightarrow \sin \frac{{B - A}}{2}(tgA - tgB) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin \frac{{B - A}}{2} = 0\\ tgA = tgB \end{array} \right. \Leftrightarrow A = B \end{array}$
Ta có

$DPCM$

$3/$

$a + b = tg\frac{C}{2}(atgA + btgB)$

$\begin{array}{l} \Leftrightarrow (a + b)\cot g\frac{C}{2} = atgA + btgB\\ \Leftrightarrow a(\cot g\frac{C}{2} - tgA) = b(tgB - \cot g\frac{C}{2}) \end{array}$
Vậy

$3/$ chính là dạng tương đương của

$2/$

$4/$

$tgA + tgB = 2\cot g\frac{C}{2} \Leftrightarrow \frac{{\sin (A + B)}}{{\cos A\cos B}} = 2\frac{{c{\rm{os}}\frac{C}{2}}}{{\sin \frac{C}{2}}}$

$\Leftrightarrow \frac{{2\sin \frac{C}{2}c{\rm{os}}\frac{C}{2}}}{{\cos A\cos B}} = 2\frac{{c{\rm{os}}\frac{C}{2}}}{{\sin \frac{C}{2}}} (1)$
Do

$c{\rm{os}}\frac{C}{2}\# 0$ ,nên (

$1$ ) tương đương:

$2{\sin ^2}\frac{C}{2} = 2\cos A\cos B$

$\begin{array}{l} \Leftrightarrow 1 - \cos C = c{\rm{os}}(A + B) + c{\rm{os}}(A - B)\\ \Leftrightarrow 1 - \cos C = - \cos C + c{\rm{os}}(A - B)\\ \Leftrightarrow 1 = c{\rm{os}}(A - B)\\ \Leftrightarrow A = B \end{array}$
Ta có

$DPCM$

$5/$

$\frac{1}{2}(tgA + tgB) = \frac{{\sin A + \sin B}}{{\cos A + \cos B}}$

$\Leftrightarrow \frac{{\sin (A + B)}}{{2\cos A\cos B}} = \frac{{2\sin \frac{{A + B}}{2}c{\rm{os}}\frac{{A - B}}{2}}}{{2\cos \frac{{A + B}}{2}c{\rm{os}}\frac{{A - B}}{2}}}$

$\Leftrightarrow \frac{{2\sin \frac{{A + B}}{2}c{\rm{os}}\frac{{A + B}}{2}}}{{2\cos A\cos B}} = \frac{{\sin \frac{{A + B}}{2}}}{{c{\rm{os}}\frac{{A + B}}{2}}} (*)$
Do

$\sin \frac{{A + B}}{2} = c{\rm{os}}\frac{C}{2}\# 0$ ,nên

$(*)$ tương đương

$2{\sin ^2}\frac{C}{2} = 2\cos A\cos B$

$\begin{array}{l} \Leftrightarrow 1 - \cos C = c{\rm{os}}(A + B) + c{\rm{os}}(A - B)\\ \Leftrightarrow 1 = c{\rm{os}}(A - B)\\ \Leftrightarrow A = B \end{array}$
Ta có

$DPCM$

$6/$

$\sin A\sin B = c{\rm{o}}{{\rm{s}}^2}\frac{C}{2} \Leftrightarrow c{\rm{os}}(A - B) - c{\rm{os}}(A + B) = 1 + \cos C$

$\begin{array}{l} \Leftrightarrow c{\rm{os}}(A - B) + \cos C = 1 + \cos C\\ \Leftrightarrow c{\rm{os}}(A - B) = 1\\ \Leftrightarrow A = B \end{array}$
Ta có

$ĐPCM$

$7/$

$\frac{{\sin A + \sin B + \sin C}}{{\sin A + \sin B - \sin C}} = \cot g\frac{A}{2}\cot g\frac{C}{2}$
Áp dụng các công thức đã biết

$\begin{array}{l} \sin A + \sin B + \sin C = 4\cos \frac{A}{2}c{\rm{os}}\frac{B}{2}c{\rm{os}}\frac{C}{2}\\ \sin A + \sin B - \sin C = 4\sin \frac{A}{2}\sin \frac{B}{2}c{\rm{os}}\frac{C}{2} \end{array}$
Hệ thức đã cho tương đương

$\frac{{4\cos \frac{A}{2}c{\rm{os}}\frac{B}{2}c{\rm{os}}\frac{C}{2}}}{{4\sin \frac{A}{2}\sin \frac{B}{2}c{\rm{os}}\frac{C}{2}}} = \cot g\frac{A}{2}\cot g\frac{C}{2}$

$\begin{array}{l} \Leftrightarrow \cot g\frac{A}{2}\cot g\frac{B}{2} = \cot g\frac{A}{2}\cot g\frac{C}{2}\\ \Leftrightarrow \cot g\frac{B}{2} = \cot g\frac{C}{2}\\ \Leftrightarrow B = C \end{array}$
Ta có

$ĐPCM$

$8/$ do

$c{\rm{o}}{{\rm{s}}^3}\frac{A}{2}c{\rm{o}}{{\rm{s}}^3}\frac{B}{2}\# 0$ ,nên ta có:

$\sin \frac{A}{2}c{\rm{o}}{{\rm{s}}^3}\frac{B}{2} = \sin \frac{B}{2}c{\rm{o}}{{\rm{s}}^3}\frac{A}{2}$

$\begin{array}{l} \Leftrightarrow \frac{{\sin \frac{A}{2}c{\rm{o}}{{\rm{s}}^3}\frac{B}{2}}}{{c{\rm{o}}{{\rm{s}}^3}\frac{A}{2}c{\rm{o}}{{\rm{s}}^3}\frac{B}{2}}} = \frac{{\sin \frac{B}{2}c{\rm{o}}{{\rm{s}}^3}\frac{A}{2}}}{{c{\rm{o}}{{\rm{s}}^3}\frac{A}{2}c{\rm{o}}{{\rm{s}}^3}\frac{B}{2}}}\\ \Leftrightarrow tg\frac{A}{2}(1 + t{g^2}\frac{A}{2}) = tg\frac{B}{2}(1 + t{g^3}\frac{B}{2})\\ \Leftrightarrow (tg\frac{A}{2} - tg\frac{B}{2}) + (t{g^3}\frac{A}{2} - t{g^3}\frac{B}{2}) = 0\\ \Leftrightarrow (tg\frac{A}{2} - tg\frac{B}{2})(1 + t{g^2}\frac{A}{2} + tg\frac{A}{2}tg\frac{B}{2} + t{g^2}\frac{B}{2}) = 0\\ \Leftrightarrow tg\frac{A}{2} - tg\frac{B}{2} = 0({x^2} + xy + {y^2} \ge 0)\\ \Leftrightarrow A = B \end{array}$
Ta có

$ĐPCM$

$9/$ áp dụng hệ thức

${\sin ^2}\frac{A}{2} = \frac{{1 - \cos A}}{2} = \frac{{1 - \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}}}{2} = \frac{{{a^2} - {{(b - c)}^2}}}{{4bc}}$
Hệ thức đã cho tương đương:

${\sin ^2}\frac{A}{2} = \frac{{{a^2}}}{{4bc}}$

$\begin{array}{l} \Leftrightarrow \frac{{{a^2} - {{(b - c)}^2}}}{{4bc}} = \frac{{{a^2}}}{{4bc}}\\ \Leftrightarrow {(b - c)^2} = 0\\ \Leftrightarrow b = c \end{array}$
Ta có

$ĐPCM$

$10/$ Áp dụng hệ thức

$c{\rm{o}}{{\rm{s}}^2}\frac{A}{2} = \frac{{1 + \cos A}}{2} = \frac{{1 + \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}}}{2} = \frac{{{{(b + c)}^2} - {a^2}}}{{4bc}}$

$\Rightarrow t{g^2}\frac{A}{2} = \frac{{{a^2} - {{(b - c)}^2}}}{{{{(b + c)}^2} - {a^2}}}$
Vì thế hệ thức đã cho đưa về dạng tương đương sau:

$t{g^2}\frac{A}{2} = \frac{{{a^2}}}{{(a + b + c)(b + c - a)}}$

$\begin{array}{l} \Leftrightarrow \frac{{{a^2} - {{(b - c)}^2}}}{{{{(b + c)}^2} - {a^2}}} = \frac{{{a^2}}}{{(a + b + c)(b + c - a)}}\\ \Leftrightarrow \frac{{{a^2} - {{(b - c)}^2}}}{{(a + b + c)(b + c - a)}} = \frac{{{a^2}}}{{(a + b + c)(b + c - a)}}\\ \Leftrightarrow b = c \end{array}$
Ta có

$ĐPCM$

$11/$ áp dụng công thức

${l_a} = \frac{{2bc\cos \frac{A}{2}}}{{b + c}} \Rightarrow l_a^2 = \frac{{4{b^2}{c^2}}}{{{{(b + c)}^2}}}c{\rm{o}}{{\rm{s}}^2}\frac{A}{2} = \frac{{4{b^2}{c^2}}}{{{{(b + c)}^2}}}(1 + \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}) \\= \frac{{bc}}{{{{(b + c)}^2}}}\left[ {{{(b + c)}^2} - {a^2}} \right]$
Từ đó suy ra

$l_a^2 = \frac{{4bc - {a^2}}}{4}$

$\begin{array}{l} \Leftrightarrow \frac{{bc\left[ {{{(b + c)}^2} - {a^2}} \right]}}{{{{(b + c)}^2}}} = \frac{{4bc - {a^2}}}{4}\\ \Leftrightarrow 4bc{(b + c)^2} - 4bc{a^2} = 4bc{(b + c)^2} - {a^2}{(b + c)^2}\\ \Leftrightarrow 4bc = {(b + c)^2}\\ \Leftrightarrow {(b - c)^2} = 0\\ \Leftrightarrow b = c \end{array}$
Ta có

$ĐPCM$

$12/$ ta có

$\begin{array}{l} {a^2} + {b^2} = tg\frac{C}{2}({a^2}tgA + {b^2}tgB)\\ \end{array}$

$\Leftrightarrow {a^2}(1 - tg\frac{C}{2}tgA) + {b^2}(1 - tg\frac{C}{2}tgB) = 0$

$\Leftrightarrow \frac{{{a^2}}}{{\cos A}}c{\rm{os}}(\frac{C}{2} + A) + \frac{{{b^2}}}{{\cos B}}c{\rm{os}}(\frac{C}{2} + B) = 0 (1)$
Vì

$(\frac{C}{2} + A) + (\frac{C}{2} + B) = \pi \Rightarrow c{\rm{os}}(\frac{C}{2} + B) = - c{\rm{os}}(\frac{C}{2} + A)$
Do đó

$(1) \Leftrightarrow c{\rm{os}}(\frac{C}{2} + A)({\sin ^2}A\cos B - {\sin ^2}B\cos A) = 0$

$\Leftrightarrow c{\rm{os}}(\frac{C}{2} + A)\left[ {(1 - c{\rm{o}}{{\rm{s}}^2}A)\cos B - (1 - c{\rm{o}}{{\rm{s}}^2}B)\cos A} \right] = 0$

$\Leftrightarrow c{\rm{os}}(\frac{C}{2} + A)(\cos B - \cos A)(1 + \cos A\cos B) = 0 (2)$
Do

$1+cosAcosB>$ ,nên :

$(2) \Leftrightarrow \left[ \begin{array}{l} c{\rm{os}}(\frac{C}{2} + A) = 0\\ \cos A = \cos B \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \frac{C}{2} + A = \frac{A}{2} + \frac{B}{2} + \frac{C}{2}\\ A = B \end{array} \right. \Leftrightarrow A = B$
Ta có

$ĐPCM$

$13/$ ta có