In a rectangular trapezoid ABCD, the angle ABC is straight, AD = 1, DC = root of 2, BC = 2

In a rectangular trapezoid ABCD, the angle ABC is straight, AD = 1, DC = root of 2, BC = 2. Is the statement true? 1) The diagonals of the trapezoid ABCD are divided by the point of intersection in the ratio 2: 1 2) Line AB touches the circle circumscribed about the triangle BDC.

Let us prove that the triangles AOD and BOC are similar.

For triangles, the angle AOD = BOC as vertical angles, the angle ADC = CBO as criss-crossing angles at the intersection of parallel lines AD and BC of the secant AD. Then triangles AOD is similar to BOC in two angles.

Since AD ​​/ BC = 1/2, then DO / OB = OA / OC = 1/2. Diagonals are divided in the ratio 1/2.

In the triangle DСН we define the leg DН. DH ^ 2 = CD ^ 2 – CH ^ 2 = 2 – 1 = 1. DH = 1 cm. Then the triangle DH is isosceles. In triangle BDH, the hypotenuse BD will be equal to:

BD ^ 2 = BH ^ 2 + DH ^ 2 = 1 + 1 = 2.

ВD = √2 cm.

Then in the triangle ВDC the angles at the base ВС are equal to 45, then the angle ВDC = 90.

Since the triangle BDC is rectangular, the diameter of the circle described near it is equal to the hypotenuse BC of the triangle, then the circle touches the segment AB at point B.

Answer: Statements 1 and 2 are correct.