A point D is taken on side AB of triangle ABC so that the circle passing through points A, C and D

A point D is taken on side AB of triangle ABC so that the circle passing through points A, C and D touches line BC. Find AD if AC = 40, BC = 34, and CD = 20.

Consider triangles BDC and BCA. In triangle BDC, angle DBC and in triangle BCA, angle CBA are equal, since they are common (angle B). The angle BCD and the angle BAC are also equal, since the angle BCD is the angle between the tangent to the circle and the chord drawn through the tangency point, then it is equal to half the degree measure of the arc DC, and the angle BAC is the inscribed angle based on the same arc DC (the degree measure of the inscribed angle is equal to half the degree measure of the arc on which it rests), therefore:
angle BCD = angle BAC.
Based on the equality of the two angles of the triangles BDC and BCA, these triangles are similar in the first sign of similarity (if two angles of one triangle are respectively equal to two angles of another triangle, then the triangles are similar).
Since BDC and BCA are similar triangles, the relationship is true:
AB / BC = AC / DC;
AB / 34 = 40/20;
AB = 34 * 40/20 (in proportion);
AB = 68 cm.
By the secant and tangent theorem (if a tangent and a secant are drawn to the circle from one point, then the product of the entire secant by its outer part is equal to the square of the tangent):
BC ^ 2 = AB * BD;
34 ^ 2 = 68BD;
68BD = 1156;
BD = 1156/68;
ВD = 17 cm.
Segment AB consists of two segments:
AB = BD + AD;
17 + AD = 68;
AD = 68-17;
AD = 51 cm.
Answer: AD = 51 cm.