[Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part. Sum of the angles of a triangle. Theorem on the exterior angle. The ratio in which the bisector divides the side. Criteria for similarity

The bisector of angle $C$ of triangle $A B C$ divides side $A B$ into segments equal to $a$ and $b$ ( $a>b$ ). The tangent to the circumscribed circle of triangle $A B C$, passing through point $C$, intersects line $A B$ at point $D$. Find $C D$.

#
Triangles $D A C$ and $D C B$ are similar.

## Solution

Let $C M$ be the bisector of triangle $A B C, A M=a, B M=b, B D=x, C D=y$.

First method. By the property of the bisector of a triangle, $A C: C B = A M: M B = a: b$.

From the equality of angles $A$ and $B C D$, it follows that triangles $D A C$ and $D C B$ are similar, hence $B D: C D = C D: A D = C B: A C = b: a$, or $x / y = y / (x + a + b) = b / a$. From this system, we find that $y = a b / (a - b)$.
![](https://cdn.mathpix.com/cropped/2024_05_06_a69fc833ac69b1da7555g-19.jpg?height=388&width=1510&top_left_y=1561&top_left_x=274)

Second method. Let $\angle C = 2 \gamma, \angle B C D = \angle A = \beta$. By the theorem on the exterior angle of a triangle, $\angle C M D = \gamma + \beta = \angle D C M$, which means triangle $D C M$ is isosceles, $y = D M = B D + B M = x + b$. By the theorem on the tangent and the secant, $y^2 = x(x + a + b) = (y - b)(y + a)$, from which we get $y = a b / (a - b)$.

## Answer

$a b / (a - b)$.