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

Through the vertices A, B, C of the parallelogram ABCD with sides AB = 3 BC = 5, a circle is drawn that intersects the line BD at point E, and BE = 9. Prove BE more than BD. Find the diagonal BD.

In a triangle ABD, AB = 3 cm, AD = BC = 5 cm.

By the condition of constructing triangles, AB + AD> BD, Then BD <8 cm, and since BE = 9 cm, then BE> BD, which was required to be proved.

Let’s extend AD to the intersection with the circle at point K. ABSC is an isosceles trapezoid, SK = AB = 3 cm, and since CD = AB, the triangle CDK is isosceles, CD = SK = 3 cm.

Let BD = X cm, then DE = 9 – X cm.

BE and AK are intersecting chords, then BD * DE = AD * DK.

DK = BD * DE / AD = X * (9 – X) / 5.

In a right-angled triangle СDН:

CD ^ 2 = CD ^ 2 – DH ^ 2 = 9 – (X * (9 – X) ^ 2/100).

СD = √ (9 – (9 * X – X ^ 2) ^ 2/100) cm.

CD is the height of triangle ABD.

Then Savd = AD * CH / 2 = (5/2) * √ (9 – (9 * X – X ^ 2) ^ 2/100) cm2.

The semi-perimeter of the triangle ABD is equal to: p = (8 + X) / 2 cm.

Then, by Heron’s theorem: Savd = √ (8 + X) / 2 * ((8 + X) / 2) – X) * ((8 + X) / 2) – 5) * ((8 + X) / 2) – 3) = (1/4) * √ (8 + X) * (8 – X) * (X – 2) * (X + 2).

Then: (5/2) * √ (9 – (9 * X – X ^ 2) ^ 2/100) = (1/4) * √ (8 + X) * (8 – X) * (X – 2 ) * (X + 2).

√ (900 – (9 * X – X ^ 2) ^ 2 = √ (64 – X) ^ 2 * (X – 4) ^ 2.

(900 – (9 * X – X ^ 2) ^ 2 = (64 – X) ^ 2 * (X – 4) 2.

18 * X ^ 3 – 149 * X ^ 2 + 1156 = 0.

(9 * X – 34) * (2 * X ^ 2 – 9 * X – 34) = 0.

X1 = 34/9.

X2 ^ 3 = (9 ± √353) / 4.

(9 – √353) / 4 <0. Not suitable.

When X = (9 + √353) / 4, the angle A turns out to be obtuse, then X = BD = 34/9 = 3 (7/9) cm.

Answer: Diagonal BD is 3 (7/9) cm.