cho tam giác ABC có các cạnh thỏa mãn hệ thức 
$\frac{1+cosB}{sinB}=\frac{2a+c}{\sqrt{4a^2-c^2}}$
C/m $\triangle ABC$ cân
cm tam giác dó cân –  nguyenquangtruonghktcute 29-04-16 04:21 PM
Bình phương 2 vế:
 (1+cosB)^2/(1-cos^2B) = (2R)^2(2sinA+sinC)^2/(2R)^2[(2sinA)^2-(s...
<=>(1+cosB)/(1-cosB)=(2sinA+sinC)/(2sinA...
<=>(1+cosB)(2sinA-sinC)=(2sinA+sinC)(1-c...
<=>(1+cosB)(2sinA-sinC)-(2sinA+sinC)(1-c...
<=>4sinAcosB-2sinC=0
<=>4sinAcosB-2sin(A+B)=0
<=>4sinAcosB-2(sinAcosB+sinBcosA)=0
<=>2(sinAcosB-sinBcosA)=0
<=>2sin(A-B)=0(*)
Do A,B,C thuộc (0,pi),nên từ (*) =>A-B=0
<=>A = B
Tam giác ABC cân tại C
lại copy @_@ –  nguyenquangtruonghktcute 01-05-16 10:16 AM

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