AC=10 OD-?

Triangle ABC is a right-angled triangle at C. The circle with center O is inscribed in the triangle, touching sides BC, AC, and AB at points D, K, and A respectively. This means the circle is the incircle of triangle ABC. The line segment OD is a radius of the incircle. The length AC is given as 10. The length BC is given as 24. Since the circle is tangent to AC at K and to BC at D, CKOD is a square, so CK = CD = OD = OK. Let r be the radius OD. Then CK = CD = r. AK = AC - CK = 10 - r. BD = BC - CD = 24 - r. By the Pythagorean theorem in triangle ABC, $$AB^2 = AC^2 + BC^2 = 10^2 + 24^2 = 100 + 576 = 676$$. So, AB = $$\sqrt{676} = 26$$. The lengths of tangents from a vertex to the incircle are equal. Thus, AK = AD = 10 - r and BD = BE = 24 - r. Also, AB = AD + DB = (10 - r) + (24 - r) = 34 - 2r. We know AB = 26. So, $$26 = 34 - 2r$$. This gives $$2r = 34 - 26 = 8$$, so $$r = 4$$. Therefore, OD = 4.