In trapezoid ABCD with bases AD and BC, the diagonals meet at point O. BO = 4 cm, OD = 8 cm

| August 1, 2021 | education

In trapezoid ABCD with bases AD and BC, the diagonals meet at point O. BO = 4 cm, OD = 8 cm, AC = 15 cm. Find the lengths OC and AO.

Let us prove the similarity of the triangles BOC and AOD.

The BOC angle is equal to the AOD angle as the vertical angles at the intersection of the AC and BD diagonals. Angle ОВС = ОDА as criss-crossing angles at the intersection of parallel straight lines ВС and АD of secant ВD.

Then the triangles BOC and AOD are similar in two angles.

The coefficient of similarity of triangles is: K = OB / OD = 4/8 = 1/2.

Let the length of the segment OC = X cm, then OA = (AC – X) = (15 – X) cm.

Then OC / OA = X / (15 – X) = 1/2.

2 * X = 15 – X.

3 * X = 15.

X = OC = 15/3 = 5 cm.

ОА = 15 – 5 = 10 cm.

Answer: The length of the OC is 5 cm, the AO is 10 cm.



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