In trapezoid ABCD with bases AD = 12 cm and BC = 4 cm, point P is the middle of AD.

In trapezoid ABCD with bases AD = 12 cm and BC = 4 cm, point P is the middle of AD. Diagonal BD meets line segment CP at point N. Find CN: NP

Since ABCD is a trapezoid, then AD is parallel to BC, then the angle BCH = DPH as criss-crossing angles at the intersection of parallel AD and BC secant CP. Angle BCH = DHP as vertical angles, then triangle BCH is similar to triangle DHP in two angles.

Since point P is the middle of AD, then DP = AD / 2 = 12/2 = 6 cm.

Then: CH / HP = BC / DP = 4/6 = 2/3.

Answer: CH / HP = 2/3.