[ Pythagorean Theorem (direct and inverse) ] [Auxiliary similar triangles]

In a right triangle $ABC$, $AC=16$, $BC=12$. A circle is described from the center $B$ with radius $BC$, and a tangent is drawn to it, parallel to the hypotenuse $AB$ (the tangent and the triangle lie on opposite sides of the hypotenuse). The leg $BC$ is extended to intersect with the drawn tangent. Determine how much the leg is extended.
Consider similar triangles.

## Solution

Let $M$ be the point of tangency, $K$ be the point of intersection of the tangent with the extension of the leg $CB$. $AB^2 = CB^2 + CA^2 = 144 + 256 = 400$.

Triangles $BMK$ and $ACB$ are similar, so $BK : BM = AB : AC$. Therefore, $BK = 20 \cdot 16 : 12 = 15$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-15.jpg?height=440&width=481&top_left_y=2064&top_left_x=800)

## Answer

15.