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Isosceles, inscribed, and circumscribed trapezoids [Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part

An isosceles trapezoid with bases $A D$ and $B C (A D > B C)$ is circumscribed around a circle, which touches the side $C D$ at point $M$. The segment $A M$ intersects the circle at point $N$. Find the ratio of $A D$ to $B C$, if $A N: N M = k$.

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## First solution.

Let $A D=a, B C=b, M N=x$. Let $Q$ and $F$ be the points of tangency of the circle with sides $B C$ and $A B$ respectively (Fig.1). Then, by the tangent-secant theorem,

$$
A F^2 = A M \cdot A N, \text{ and } \frac{t^2}{t} = k(k+1) \times 2
$$

Extend $A M$ to intersect line $B C$ at point $P$. From the similarity of triangles $C P M$ and $D A M$, it follows that $\frac{C P}{C M} = \frac{A D}{D M} = 2$. Therefore, $C P = 2 C M = 2 C Q = b$.

By the tangent-secant theorem,

$$
P Q^2 = P N \cdot P M, \text{ and } \frac{9 b^2}{t} = \left(\frac{b x(k+1)}{a} + x\right) \cdot \frac{b x(k+1)}{a}
$$

(since $P M = \frac{b}{a} \cdot A M$). Dividing this equality term by term by the previously proven one, we get $9 b k = b(k+1) + a$.

From this, we find that $\frac{d}{b} = 8 k - 1$.

## Second solution.

Let $A D=a, B C=b, M N=x$ (Fig.2). Let $F$ and $L$ be the points of tangency of the circle with sides $A B$ and $A D$ respectively. Then $M F = \frac{2 a t b}{a+b}, A L = \frac{a}{2}$.

We will prove that line $N F$ bisects segment $A L$. Indeed, let lines $N F$ and $A D$ intersect at point $K$. From the parallelism of lines $F M$ and $A D$ and the theorem about the angle between a tangent and a chord, it follows that

$$
\angle N A K = \angle F M N = \angle A F K
$$

which means that triangles $A F K$ and $N A K$ are similar by two angles, so $\frac{K A}{K F} = \frac{K N}{K A}$, hence $K A^2 = K F \cdot K N$. On the other hand, by the tangent-secant theorem, $K L^2 = K F \cdot K N$. Therefore, $K A = K L$, which is what we needed to prove.

From the similarity of triangles $A N K$ and $M N F$, it follows that $\frac{A K}{M I} = \frac{A N}{M N}$, or

$$
\frac{\frac{a}{4}}{\frac{2 a b}{a+b}} = k \Leftrightarrow \frac{a+b}{8 b} = k \Leftrightarrow \frac{a}{b} + 1 = 8 k \Leftrightarrow \frac{a}{b} = 8 k - 1
$$

## Answer

$8 k - 1$.