Two circles touch each other internally at point A. The smaller circle touches the chord BC

Two circles touch each other internally at point A. The smaller circle touches the chord BC of the larger circle at point D. It is known that AB = 24, AC = 40, AD = 15. Find the radius of the larger circle.

Designations
AB = c;
AC = b = 40;
BC = a = 24;
CH = l = 15;
∠A = α;
∠B = β;
∠C = γ;
BH = x;
AH = y;
R, r are the radii of the circles.

CH is the bisector of triangle ABC
1. Draw the chord BD ⊥ O1O2. The radius perpendicular to the chord divides the arc contracted by this chord in half:

∠CBD = ∠BAC = α;

∠ABD = ∠ABC – ∠CBD = β – α.

2. O1H ⊥ AB, therefore:

∠HO1O2 = ∠ABD = β – α.

In an isosceles triangle HCO1 – external angle ∠HO1O2:

∠O1HC = ∠O1CH = 1/2 * ∠HO1O2 = 1/2 * (β – α).

3.∠BPC = 90 °:

∠BCP = 90 ° – ∠CBP = 90 ° – α;

∠BCH = ∠BCP – ∠O1CH = 90 ° – 1/2 * (β + α).

4. The angle BCH is equal to half of ∠ACB:

1/2 * ∠ACB = 1/2 * (180 ° – ∠ABC – ∠BAC) = 1/2 * (180 ° – β – α) = 90 ° – 1/2 * (β + α) = ∠BCH ,

hence CH is the bisector of triangle ABC.

Property of the bisector of a triangle
CH divides side AB into parts proportional to sides BC and AC:

BH: ВС = AH: AC = k, where k is the proportionality coefficient;

x: a = k; => x = ak;

y: b = k; => y = bk.

Let’s apply the cosine theorem for BCH and ACH:

{x² = a² + l² – 2al * cos (γ / 2)
{y² = b² + l² – 2bl * cos (γ / 2)

{2al * cos (γ / 2) = a² (1 – k²) + l²
{2bl * cos (γ / 2) = b² (1 – k²) + l²

Having solved the system of equations, we get:

k = √ (1 – l² / ab);

k = √ (1 – 15² / 24 * 40) = 7/8;
x = ka = 7/8 * 24 = 21;
y = kb = 7/8 * 40 = 35;
c = x + y = 21 + 35 = 56.
According to the well-known formula for the radius of the circumscribed circle:

R = 1/4 * abc / √ (p * (p – a) * (p – b) * (p – c)),

where p = 1/2 * (a + b + c) is the semiperimeter of the triangle, we calculate R:

R = 56 / √3;

Answer: 56 / √3.