When you draw an altitude from the right angle to the hypotenuse in a right triangle, it creates two smaller right triangles that are similar to the original triangle. Use this idea to relate a leg (like $BC$) to the hypotenuse and one of its segments.

$BC$ is the leg that touches the segment $DB$ of the hypotenuse. There is a relationship of the form $\text{(leg)}^2 = \text{(adjacent segment)} \times \text{(hypotenuse)}$. Apply this to $BC$, $DB$, and $AB$.

Once you have an equation for $BC^2$ in terms of $DB$ and $AB$, plug in the numerical values and then take the square root, simplifying the radical if possible.

In Desmos, type 18*26 to calculate the value of $BC^2 = DB \cdot AB$. Note the numerical result that appears.

In a new Desmos line, type sqrt(18*26) (or sqrt( followed by the result from the previous step). Use that output as the length of $BC$, and then, if needed, factor the number under the square root to see which perfect square can be taken out to simplify the radical.

We have right triangle $\triangle ABC$ with a right angle at $C$. The altitude from $C$ meets hypotenuse $\overline{AB}$ at $D$, splitting $AB$ into $AD = 8$ and $DB = 18$.

So the full hypotenuse length is

When the altitude is drawn from the right angle to the hypotenuse in a right triangle, it creates two smaller right triangles that are each similar to the original triangle.

From this similarity, there is a key relationship: each leg of the original triangle is the geometric mean of the entire hypotenuse and the segment of the hypotenuse adjacent to that leg.

For leg $BC$, which is adjacent to segment $DB$, this gives the equation

We know $DB = 18$ and we found $AB = 26$ in Step 1. Substitute into the equation from Step 2:

Now multiply:

To find $BC$, take the square root of both sides:

Factor 468 to pull out a perfect square:

So the length of $BC$ is $6\sqrt{13}$.