In an isosceles trapezoid, the diagonals are bisectors of acute angles and are divided at the point

In an isosceles trapezoid, the diagonals are bisectors of acute angles and are divided at the point of intersection in a ratio of 13: 5, starting from the vertices of the acute angles. Find the perimeter of the trapezoid if its height is 32cm.

Since the trapezoid is isosceles, then its angles at the base are equal, and since AC and BD are bisectors of acute angles, then the angle BAC = CAD = CDB = BDA.

Angle ВСD = ВDА as cross-lying angles at the intersection of parallel ВС and АD of secant ВD.

Then the angle CBD = CDB, which means the triangle is equilateral and BC = CD, and since AB = CD, then AB = BC = CD.

The diagonals of the trapezoid form similar triangles at the bases when they intersect.

The BOC triangle is similar to the AOD. According to the condition AO / CO = 13/3, then AD / BC = 13/5.

Let AD = 13 * X, then BC = AB = CD = 13 * X.

The height BH cuts off, on the basis of AD, the segment AH equal to the half-difference of the bases.

AH = (AD – BC) / 2 = (13 * X + 5 * X) / 2 = 4 * X.

Then in a right-angled triangle ABN, by the Pythagorean theorem, BH ^ 2 = AB ^ 2 – AH ^ 2 = (5 * X) 2 – (4 * X) 2.

1024 = 9 * X2

.X = 32/3.

AB = BC = CD = 5 * 32/3 = 160/3.

AD = 13 * 32/3 = 416/3.

Ravsd = 3 * AB + AD = 160 + 416/3 = 896/3 = 298 (2/3) cm.

Answer: The perimeter of the trapezoid is 298 (2/3) cm.