The bisector of the angle BAC intersects the circle described about the triangle ABC at point D

The bisector of the angle BAC intersects the circle described about the triangle ABC at point D, so that BC = 4 DC = √5 find the radius of the circle and the cosine of the angle BAC.

From the center of the circle O we construct the radii OB, OD, OC.

We denote their length by R.

<BAD = <DAC (formed by the bisector AD).

The central angles are equal and resting on the same arcs:

<BOD = <DOC.

ΔBОС is isosceles, the segment OK is its bisector, and, consequently, the height and median.

The BC segment is perpendicular to OK.

KC = BK = BC / 2 = 2.

KD = √ (DC ^ 2 – KC ^ 2) = √ ((√5) ^ 2 – 2 ^ 2) = 1.

OC ^ 2 = KC ^ 2 + (OD – KD) ^ 2;

R ^ 2 = 2 ^ 2 + (R – 1) ^ 2

R ^ 2 = 4 + R ^ 2 – 2R +1;

2R = 5

R = 2.5.

OK = OD – KD = 2.5 -1 = 1.5;

cos (<KOC) = OK / OC = 1.5 / 2.5 = 0.6.

<KOC = 2 * <DAC (central and inscribed corners);

2 * <DAC = <BAC;

<KOC = <BAC;

cos (<BAC) = cos (<KOC) = 0.6.

Answer: R = 2.5; cos (<BAC) = 0.6.