Trapezium bases 6cm and 9cm, height 10cm. Find the distance from the point of intersection

Trapezium bases 6cm and 9cm, height 10cm. Find the distance from the point of intersection of the diagonals of the trapezoid to its bases.

Initial data: AB = 6 cm, DC = 9 cm, KH = 10 cm.
1. Since DC is parallel to AB (the base of the trapezoid), the angles BDC and DBA are equal as crosswise at the secant DB.
The angles DMC and AMB are vertical, therefore, they are equal to each other.
Thus, triangles DMC and AMB are similar in 2 angles.
2. KM and MH are the heights of triangles AMB and DMC, respectively.
KM = KH – MH = (10 – MH) cm.
KM / MH = AB / DC (property of similar triangles).
(10 – MH) / MH = 6/9 = 2/3.
3 * (10 – MH) = 2 * MH;
30 – 3 * MH = 2 * MH;
30 = 5 * MH;
MH = 6 cm.Then KM = 4 cm.
Answer: 4 cm., 6 cm.