# 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 (http://bit.ly/2AF4yXB). 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.