A circle is drawn through the vertices A, B, C of the parallelogram ABCD with sides AB = 3 and BC = 5

A circle is drawn through the vertices A, B, C of the parallelogram ABCD with sides AB = 3 and BC = 5, intersecting line BD at point E, and BE = 9. Find the diagonal BD.

Let us draw the diagonals of the parallelogram with the intersection point O.

Let the segment ОВ be equal to X cm, and the segment ОC = Y cm. Since the diagonals at the intersection point are halved, then ОВ = ОD, and ОC = ОА.

Since the sum of the squares of the diagonals of the parallelogram is equal to the sum of the squares of its sides, then: AC ^ 2 + BC ^ 2 = 2 * (AB ^ 2 + BC ^ 2) = 2 * (9 + 25) = 68.

Since AC ^ 2 = (2 * OC) ^ 2 = 4 * Y ^ 2, and

BC ^ 2 = 2 * OB ^ 2 = 2 * X ^ 2, then

4 * Y ^ 2 + 4 * X ^ 2 = 68.

Y ^ 2 + X ^ 2 = 17. (1).

AC and BE are intersecting chords at the point O, then by the intersecting chord theorem

AO * CO = BO * EO.

Since OE = 9 – X, then

Y ^ 2 = X * (9 – X).

X ^ 2 + Y ^ 2 = 9 * X. (2).

We destroy the system of equations (1) and (2).

9 * X = 17.

X = 17/9.

BD = 2 * X = 34/9 = 3 (7/9) cm.

Answer: ВD = 3 (7/9) cm.