The diagonals AC and BD of the quadrilateral ABCD intersect at the point О, AO = 18 cm, ОВ = 15 cm

| December 12, 2020 | education

The diagonals AC and BD of the quadrilateral ABCD intersect at the point О, AO = 18 cm, ОВ = 15 cm, OC = 12 cm, OD = 10 cm. Prove that ABCD is a trapezoid

Consider triangles AOB and COD, in which the angle AOB = COD as the vertical angles at the intersection of the diagonals AC and BD.

OD / CO = 10/12 = 5/6.

ОВ / ОА = 15/18 = 5/6.

Since the ratio of the lengths of the sides of the triangles are equal, these sides are proportional, and then the AOB triangle is similar to the COD triangle in two proportional sides and the angle between them.

Then the angle OAB = OCD, and since these are cross-lying angles at the intersection of lines AB and CD of the secant AC, then AB is parallel to CD, which means that the quadrilateral ABCD is a trapezoid, which was required to prove.



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