The case formula for a secant?

Question

There is a 3-4-5 right triangle with the hypotenuse as the base. The altitude is drawn from the hypotenuse to the right angle. A circle is then drawn with the diameter being the altitude. What is the distance from the point where the vertex of 3 and 5 is to the point where the circle intersects the side of length 3?

Answer 1

The part of a secant outside a circle times the whole secant is equal to the part of another secant that's outside times that whole secant and also is equal to the square of a tangent segment drawn from the same outside point.

By similar triangles, the tangent segment drawn from the 3/5 vertex is 9/5. Squared, that's 81/25

So 81/25 = 3x where x is the piece you need. Solve this for x.

(edited, typo first time)

Answer 2

Triangle ABC
AB = 3, AC = 4, BC = 5
Perpendicular from A to BC is AD
Circle cuts AB at P
BP =?

Triangle area = base x height / 2

5 x AD/2 = 3 x 4 /2
AD = 12/5

Triangle ADB
AD^2 + BD^2 = AB^2
BD = sqrt(AB^2 – AD^2) = 1.8

Tangent and secant relation in a circle
(BP)( AB) = BD^2
(BP) (3) = (1.8)^2 when BD = 1.8
BP = (1.8^2)/3

BP = 1.08