# On the side AB of triangle ABC, a point D is taken so that the circle passing through points A, C, and D touches the line BC

**On the side AB of triangle ABC, a point D is taken so that the circle passing through points A, C, and D touches the line BC. Find AD if AC = 40, BC = 45, and CD = 24**

The straight line and the circle can be located relative to each other in three versions:

1) Do not intersect, that is, do not have a single common point.

2) Touch, that is, have only one common point, then the line is called the tangent to the circle.

3) Intersect, that is, have two common points.

The condition of the problem says that the circle passes through point C and touches the straight line BC. This means that the line BC cannot, besides the point of tangency of other common points with the circle, have a circle, therefore, the circle touches the line BC at point C (as shown in the figure).

Consider the triangles ABC and CDB.

∠B – common

∠DAC is inscribed in a circle and rests on the CD arc. Those. equal to half its degree measure.

∠BCD encloses the CD arc as a tangent to the chord and is also equal to half the degree measure of the CD arc (in the fourth angle property).

Therefore, the angles of DAC and BCD are equal.

Then, according to the first sign of the similarity of triangles, these triangles are similar.

Hence:

AC / CD = BC / BD = AB / BC

AC / CD = BC / BD

40/24 = 45 / BD => BD = 24 * 45/40 = 27

BC / BD = AB / BC

45/27 = AB / 45 => AB = 45 * 45/27 = 75

AD = AB-BD = 75-27 = 48

Answer: AD = 48