见解析
在△ ABC 中, a cos 2
+ c ·cos 2
=
⇔
⇔ a (1+cos C )+ c (1+cos A )=3 b ⇔ a + c + a cos C + c cos A =3 b
⇔ a + c + a
=3 b ⇔ a + c +
=3 b
⇔ a + c + b =3 b ⇔ a + c =2 b ⇔ a , b , c 成等差数列.所以命题成立.
见解析
在△ ABC 中, a cos 2
+ c ·cos 2
=
⇔
⇔ a (1+cos C )+ c (1+cos A )=3 b ⇔ a + c + a cos C + c cos A =3 b
⇔ a + c + a
=3 b ⇔ a + c +
=3 b
⇔ a + c + b =3 b ⇔ a + c =2 b ⇔ a , b , c 成等差数列.所以命题成立.