10.4. The incircle of triangle $A B C$ touches side $A C$ at point $D$. A second circle passes through point $D$, touches ray $B A$ at point $A$, and also touches the extension of side $B C$ beyond point $C$. Find the ratio $A D: D C$.
Answer: $3: 1$.

Solution. Let the second circle touch the ray $B C$ at point $K$, and the inscribed circle touch the sides $A B$ and $B C$ of the triangle at points $P$ and $Q$ respectively (see Fig. 10.4). We can reason in different ways.

First method. Let $C D = C Q = x$ and $A D = A P = k x$. Since $B A = B K$ and $B P = B Q$, it follows that $P A = Q K$, hence $C K = k x - x$. By the tangent-secant theorem: $C D \cdot C A = C K^{2}$, that is, $x(x + k x) = (k x - x)^{2}$. Dividing both sides of this equation by $x^{2}$, expanding the brackets, and combining like terms, we get: $k^{2} = 3 k$, hence $k = 3$.

Second method. Let $B C = a, A C = b, A B = c$, then $C D = \frac{a + b + c}{2} - c = \frac{a + b - c}{2}, C K = B K - B C = A B - B C = c - a$. By the tangent-secant theorem: $C D \cdot C A = C K^{2}$, that is, $\frac{a + b - c}{2} \cdot b = (c - a)^{2} \Leftrightarrow b^{2} - (c - a) b - 2(c - a)^{2} = 0$. Solving this equation as a quadratic in $b$, we get: $b = a - c$ or $b = 2(c - a)$.

The equality $b = a - c$ contradicts the triangle inequality, while if $b = 2(c - a)$, then $\frac{A D}{D C} = \frac{b + c - a}{2} : \frac{a + b - c}{2} = \frac{3 c - 3 a}{c - a} = 3$.

Grading criteria.

“+” A complete and well-reasoned solution is provided

“±” A generally correct reasoning is provided, containing minor gaps or inaccuracies, and the correct answer is obtained

“±” The equation is correctly and reasonably formulated, but a computational error is made in solving it

“-” Only the answer is provided

“-” The problem is not solved or is solved incorrectly