$7 \cdot 29$ In a right-angled triangle, one side is the mean proportional between the other two sides, and the trigonometric function value of the smallest interior angle $\alpha$ equals $\frac{\sqrt{5}-1}{2}$. Then this trigonometric function is
(A) $\operatorname{ctg} \alpha$.
(B) $\operatorname{tg} \alpha$.
(C) $\cos \alpha$.
(D) $\sin \alpha$.
(Anhui Province, Anqing and other counties and cities Junior High School Mathematics Competition, 1991)
[Solution 1] In the right triangle $\triangle ABC$, the altitude $AD$ on the hypotenuse is the mean proportional between $BD$ and $DC$ (as shown in the figure).

If one of the legs of the triangle is equal to the length of $BD$ or $DC$, without loss of generality, let $AC = BD$. Then $\mathrm{Rt} \triangle ADC$ is the triangle that meets the problem's conditions.
Let $BD = a$, $DC = b$, then $AC = a$, and
$$
AD = x = \sqrt{ab},
$$

At the same time, $AB = \sqrt{a^2 + x^2} = \sqrt{a^2 + ab}$.
In $\mathrm{Rt} \triangle ABC$, by the Pythagorean theorem, we have
$$
\left(\sqrt{a^2 + ab}\right)^2 + a^2 = (a + b)^2,
$$

which simplifies to
$$
b^2 + ab - a^2 = 0, \text{ or } \left(\frac{b}{a}\right)^2 + \left(\frac{b}{a}\right) - 1 = 0,
$$

Solving this, we get
$$
\frac{b}{a} = \frac{-1 \pm \sqrt{5}}{2} \text{ (discard the negative value). }
$$

Noting that $a > b$, and $x = \sqrt{ab} > \sqrt{b^2} = b$, we know that $b$ is the shortest, so its opposite angle is the smallest.
In $\mathrm{Rt} \triangle ADC$, $\frac{b}{a} = \sin \angle CAD$,

which means
$$
\sin \angle CAD = \frac{\sqrt{5} - 1}{2}.
$$

Therefore, the answer is $(D)$.
[Solution 2] In the right triangle $\triangle ABC$, $\angle C = 90^\circ$, and let $\angle A < \angle B$. Let $a, b, c$ be the sides opposite to $\angle A, \angle B, \angle C$ respectively. According to the problem, we have
$$
b^2 = ac,
$$

and $a^2 + b^2 = c^2$, so
$c^2 - a^2 = ac$, or $\left(\frac{a}{c}\right)^2 + \frac{a}{c} - 1 = 0$.
Solving this, we get $\frac{a}{c} = \frac{\sqrt{5} - 1}{2}$ (discarding the negative value).
By definition, $\frac{a}{c} = \sin A$, so the sine of the smallest angle $\angle A$ is $\frac{\sqrt{5} - 1}{2}$. Therefore, the answer is $(D)$.