Two sides of a triangle are equal to 3 and 5. It is known that the circle passing through the midpoints

Two sides of a triangle are equal to 3 and 5. It is known that the circle passing through the midpoints of these sides and their common vertex touches the third side of the triangle. Find the third side.

Let AB = 3, BC = 5, side AC – tangent.
The circle intersects side AB at point M, which is the middle of AB, then:
AM = MB = AB / 2;
AM = 3/2.
The circle intersects the side BC at point N, which is the middle of AC, then:
CN = NB = BC / 2;
CN = 5/2.
The circle touches side AC at point D.
The tangent and secant theorem says that if a tangent and a secant are drawn from one point to a circle, then the product of the entire secant by its outer part is equal to the square of the tangent. Then:
AD ^ 2 = AB * AM;
AD ^ 2 = 3 * (3/2) = 9/2;
AD = √ (9/2) = 3 / √2 = 3√2 / 2.
And:
CD ^ 2 = CB * CN;
CD ^ 2 = 5 * (5/2) = 25/2;
CD = √ (25/2) = 5 / √2 = 5√2 / 2.
The AC side is equal to:
AC = AD + CD;
AC = 3√2 / 2 + 5√2 / 2 = (3√2 + 5√2) / 2 = 8√2 / 2 = 4√2.
Answer: AC = 4√2.