7. Let $A B C$ be a right triangle at $B$. Let $H$ be the foot of the altitude from $B$ and let $D$ be the intersection of the angle bisector of $\angle A$ with side $B C$. Suppose that $H D$ is perpendicular to $B C$. What is $\left(\frac{A B}{B C}\right)^{2}$?
(A) $\frac{\sqrt{5}}{2}$
(B) $\frac{\sqrt{5}-1}{2}$
(C) $\sqrt{5}$
(D) $\frac{\sqrt{5}+1}{2}$
(E) Such a configuration cannot be realized.
7. The answer is $\mathbf{( B )}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_04503179b337e2a9d877g-08.jpg?height=331&width=591&top_left_y=1773&top_left_x=778)

The triangles $B A C, H B C, H A B$ are similar, hence they provide the equalities

$$
\frac{A B}{B C}=\frac{B H}{C H}=\frac{A H}{B H}
$$

The lines $D H$ and $A B$ are parallel, because both are perpendicular to $B C$; by Thales' theorem and the angle bisector theorem, we have that

$$
\frac{A H}{C H}=\frac{B D}{C D}=\frac{A B}{A C}
$$

Now we can combine all our observations to obtain a relationship between the sides of the triangle:

$$
\left(\frac{A B}{B C}\right)^{2}=\frac{A B}{B C} \cdot \frac{A B}{B C}=\frac{B H}{C H} \cdot \frac{A H}{B H}=\frac{A H}{C H}=\frac{A B}{A C},
$$

which we rewrite as $A C / B C=B C / A B$. Substituting this into the Pythagorean theorem

$$
1=\frac{A C^{2}}{B C^{2}}-\frac{A B^{2}}{B C^{2}}=\left(\frac{B C}{A B}\right)^{2}-\left(\frac{A B}{B C}\right)^{2} .
$$

The ratio we are looking for is thus the positive solution of $x^{2}+x-1$.