To solve this problem, we can use a property of tangents from a point to a circle. If two tangents are drawn to a circle from an external point, the lengths of those tangents are equal.

In the diagram, we have a triangle with all sides tangent to the circle. Label the triangle vertices as $A$, $B$, and $C$, where:

• $AB = 13$,
• $BC = 7$,
• $CA = 8$.

Now, let's denote the tangents from points $A$, $B$, and $C$ as follows:

• Let the tangents from $A$ to the circle be $x$ and another length.
• Let the tangents from $B$ to the circle be $y$ and $z$.
• Let the tangents from $C$ to the circle be $z$ and $x$.

Thus, the relationships between the tangents from each vertex are:

1. $AB = x + y = 13$,
2. $BC = y + z = 7$,
3. $CA = z + x = 8$.

We now have a system of three equations:

$x + y = 13,$ $y + z = 7,$ $z + x = 8.$

Solving the system:

1. From $x + y = 13$, we get $y = 13 - x$.
2. Substitute $y = 13 - x$ into $y + z = 7$: $(13 - x) + z = 7 \implies z = x - 6.$
3. Substitute $z = x - 6$ into $z + x = 8$: $(x - 6) + x = 8 \implies 2x - 6 = 8 \implies 2x = 14 \implies x = 7.$

Thus, the measure of the segment labeled $x$ is $\boxed{7}$.

Would you like any further explanation, or do you have questions about specific steps?

Related Questions:

1. What is the length of the other tangent from point $A$?
2. How do you find the radius of the inscribed circle in such a triangle?
3. What formula would you use to find the area of the triangle if you know the side lengths?
4. How do the properties of tangents relate to the triangle’s perimeter?
5. How could you use the semiperimeter to solve for the tangents?

Tip:

In geometry, the properties of tangents are particularly useful in problems involving circles inscribed in polygons, as the tangents drawn from external points are always equal in length.