Diagonals AC and BD of quadrilateral ABCD meet at point O, AO = 18, OB = 15cm, OC = 12cm

Diagonals AC and BD of quadrilateral ABCD meet at point O, AO = 18, OB = 15cm, OC = 12cm, OD = 10, prove that ABCD is a trapezoid.

In triangles AOB and COD, the angle AOB is equal to COD as the vertical angles at the intersection of the diagonals AC and BD.

Let’s define the ratio of the segments ОD / СО and ОВ / ОА.

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

ОВ / ОА = 15/18 = 5/6.

Since the ratio of the sides is 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 ОDС = ОВА, and since these are criss-crossing angles, then AB is parallel to CD, and then ABCD is a trapezoid, which was required to be proved.