In trapezoid ABCD with bases AD and BC, the diagonals intersect at point O, BO: OD = 2: 3

| August 16, 2021 | education

In trapezoid ABCD with bases AD and BC, the diagonals intersect at point O, BO: OD = 2: 3, AC = 25 cm. Find AO and OC.

Let us prove that triangle BOС is similar to triangle AOD.

Angle BOC = AOD as vertical angles at the intersection of diagonals BD and AC.

The secant AC forms two criss-crossing angles BCO and DAO crossing parallel BC and AD, the angle BCO = DAO, then the triangles BOC and DOA are similar in two angles, the first sign of similarity of triangles.

Let the length of the segment OС = Y cm, then OA = (25 – Y) cm, the length of the segment OB = 2 * X, then OD = 3 * X.

In such triangles: ОВ / ОD = ОС / ОА.

2 * X / 3 * X = Y / (25 – Y).

3 * Y = 2 * (25 – Y).

5 * Y = 50.

Y = OС = 50/5 = 10 cm.

AO = 25 – 10 = 15 cm.

Answer: The length of the OS segment is 10 cm, AO is 15 cm.



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