In trapezoid ABCD, BCIIAD, AB_LBD, points M and K are the midpoints of segments BC and CD

| February 20, 2021 | education

In trapezoid ABCD, BCIIAD, AB_LBD, points M and K are the midpoints of segments BC and CD, respectively, MK = √5cm, AD = 2 * √10cm: a) find the angle DBC, find BE, if CE is the height of the triangle BCD, the tangent of the angle ECD is equal to 3.

Consider the BCD triangle, in which the segment MK is the middle line, since, by condition, the points M and K are the midpoints of the segments BC and BD. Then BD = 2 * MK = 2 * √5 cm.

Consider a right-angled triangle of ABD, in which hypotenuse AD = 2 * √10, and leg BD = 2 * √5, then

Cosvda = HP / BP = 2 * √5 / 2 * √10 = 1 / √2 = √2 / 2.

Angle BDA = 450. Then the angle CBD = BDA as criss-crossing angles, and the angle CMK = MBD as the corresponding angles. DBC angle = DВA = СMK = 450.

Consider a triangle BEC, in which the angle E is straight, since CE is the height of the ВСD triangle, angle B = 450, then the angle C = 450, and therefore the triangle BEC is isosceles, BE = CE.

BE = ВD – ED = 2 * √5 – ED.

BE = CE = ED / tgesd = ED / 3.

2 * √5 – ED= ED / 3.

6 * √5 = 4 * ED

ED = 3 * √5 / 2.

BE = 2 * √5 – 3 * √5 / 2 = √5 / 2 cm.

Answer: Angle DВC = 450, BE = √5 / 2 cm.



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