135 In $\triangle A B C$, the sides opposite to angles $A, B, C$ are denoted as $a, b, c (b \neq 1)$. If $\frac{C}{A}, \frac{\sin B}{\sin A}$ are both roots of the equation $\log _{\sqrt{5}} x=\log _{b}(4 x-4)$, then $\triangle A B C$
A. is an isosceles triangle, but not a right triangle
B. is neither an isosceles triangle nor a right triangle
C. is an isosceles right triangle
D. is a right triangle, but not an isosceles triangle
135 D. From the given equation, we have
$$
\log _{b} x^{2}=\log _{b}(4 x-4),
$$

so
$$
x^{2}-4 x+4=0,
$$

the roots of which are $x_{1}=x_{2}=2$, hence $C=2 A$ and $\sin B=2 \sin A$.
Since
$$
A+B+C=180^{\circ},
$$

we have
$$
3 A+B=180^{\circ},
$$

thus
$$
\sin B=\sin 3 A .
$$

Combining $\sin B=2 \sin A$ gives
$$
\sin 3 A=2 \sin A
$$

which is
$$
3 \sin A-4 \sin ^{3} A=2 \sin A .
$$

Therefore,
$$
\sin A\left(1-4 \sin ^{2} A\right)=0,
$$

but $\sin A>0$, so
$$
\sin ^{2} A=\frac{1}{4} .
$$

Thus $\sin A=\frac{1}{2}, A=30^{\circ}, C=60^{\circ}, B=90^{\circ}$, hence the answer is B.