The bases BC and AD of trapezoid ABCD are 7 and 9, respectively. The sides are extended to the intersection

The bases BC and AD of trapezoid ABCD are 7 and 9, respectively. The sides are extended to the intersection at point K. Find the height KF of triangle ADK if the height of the trapezoid is 5.

Consider triangles KAD and KBC. In them, the angle K is common, ∠B = ∠A and ∠C = ∠D, as the corresponding angles for parallel BC and AD and secants AK and DK. We conclude that these triangles are similar in three corners. The proportionality of the sides follows from the similarity of triangles.
In the triangle КБС we denote the height КН х, then the height KF = (x + 5). We write down the ratio of linear elements:
KF / KH = AD / BC
(x + 5) / x = 9/7
9x = 7x + 35
2x = 35
x = 17.5 = KH;
KF = 17.5 + 5 = 22.5
Answer: KF is 22.5.