In an isosceles trapezoid, the bases are 1: 3, the diagonal is 42 cm. The middle of one of the lateral sides and the end of the larger base, which does not belong to this side, are connected by a segment. What parts did the trapezoid diagonal divide into?
Let the length of the smaller base of the trapezoid be 2 * X cm, then, by condition, the length of the larger base is 6 * X cm.
Let us construct the middle line of the MR of the trapezoid, then at point O it divides the diagonal AC in half, AO = OC = AC / 2 = 42/2 = 21 cm.
In the ABC triangle, the MO segment is its middle line, then MO = BC / 2 = X cm.
Let’s connect points O and D. In the formed trapezoid АСОD, MO = X cm, AD = 6 * X cm, AO = 21 cm.
Triangles MOK and AKD are similar in two angles, and the coefficient of their similarity is: K = MO / AD = 1/6.
Then OK / AK = 1/6.
AK = 6 * OK.
AK + OK = AO = 21 cm.
6 * OK + OK = 21.
OK = 21/7 = 3 cm.
Then AK = 21 – 3 = 18 cm.
CK = AC – AK = 42 – 18 = 24 cm.
Answer: The diagonal is divided into 18 cm and 24 cm segments.