7.56 If the sides of a triangle $a, b, c$ are opposite to angles $A, B, C$ respectively, and satisfy $a \cos A + b \cos B = c \cos C$, then this triangle must be
(A) An equilateral triangle.
(B) A right triangle with $a$ as the hypotenuse.
(C) A right triangle with $b$ as the hypotenuse.
(D) None of the above.
(Shanghai Junior High School Mathematics Competition, 1983)
［Solution 1］From the given and the cosine rule, we have
$$
\frac{a\left(b^{2}+c^{2}-a^{2}\right)}{2 b c}+\frac{b\left(a^{2}+c^{2}-b^{2}\right)}{2 a c}=\frac{c\left(a^{2}+b^{2}-c^{2}\right)}{2 a b},
$$

which simplifies to \(a^{2}\left(b^{2}+c^{2}-a^{2}\right)+b^{2}\left(a^{2}+c^{2}-b^{2}\right)=c^{2}\left(a^{2}+b^{2}-c^{2}\right)\), thus
\(\left(a^{2}-b^{2}\right)^{2}=c^{4}\), or \(a^{2}=b^{2} \pm c^{2}\).
If \(a^{2}=b^{2}+c^{2}\), then the triangle is a right triangle with \(a\) as the hypotenuse;
If \(a=b^{2}-c^{2}\), then the triangle is a right triangle with \(b\) as the hypotenuse.
Therefore, the answer is \((D)\).
[Solution 2] From the given equation, it is symmetric with respect to \(a, A\) and \(b, B\). If \((B)\) is correct, then \((C)\) is also correct, which is not appropriate.

If \((A)\) is correct, then \(A=B=C=60^{\circ}\), and \(a=b=c\), making the given equation \(2=1\), which is a contradiction.
Therefore, the answer is \((D)\).