In trapezoid ABCD, the extensions of the sides AB and CD meet at point P, Q is the intersection point

In trapezoid ABCD, the extensions of the sides AB and CD meet at point P, Q is the intersection point of the diagonals of this trapezoid. Find the ratio of the length of the smaller base of a given trapezoid to the length of the larger base if it is known that the area of triangle ABQ is 1/4 of the area of triangle ACP.

We use the trapezoid property: The point of intersection of the diagonals and the point of intersection of the extension of the lateral sides and the middle of the bases of the trapezoid lie on one straight line. Then the segment РМ is the median of the triangle ВРС, МQ is the median ВСQ.

The area of ​​the triangle APC is equal to the sum of the areas of the triangles ABQ, BPQ, CPQ. Since PM is the median of BPC, then Svrm = Svrs, Svmq = Smsq, and then Sars = Sawq + 2 * Srvq.

By condition, Sars = 4 * Sawq, then 4 * Sawq = Sawq + 2 * Srvq.

3 * Savq = 2 * Srvq.

Srvq / Sawq = 3/2.

Triangles PBQ and ABQ have a common height omitted from a common point Q, then the ratio of the areas of these triangles is equal to the ratio of their bases PB and AB.
BО / AB = Srvq / Sawq = 3/2.

2 * BP = 3 * AB.

AB = 2 * BP / 3.

AR = AB + BP = 2 * BP / 3 + BP = 5 * BP / 3.

AP / BP = 5/3.

Triangles APD and BPC are similar, then AD / BC = AP / BP = 5/3.

BC / BP = 3/5.

Answer: The base ratio is 3/5.