In trapezoid ABCD, side AB is perpendicular to the base of BC. The circle passes through points C and D

In trapezoid ABCD, side AB is perpendicular to the base of BC. The circle passes through points C and D and touches line AB at point E. Find the distance from point E to line CD if AD = 6, BC = 5.

Point Q is the intersection point of two lines AB and CD.
Point P is the projection of point E onto line CD.
Point F is the projection of point C onto line AD.
Let’s denote the angle СDA by the letter α, and the straight line СD by the letter x.
Since FD = AD – AC = AD – BC.
Let’s substitute the values we know. We get the expression:
6 – 5 = 1.
Therefore, cos α = FD / DC = 1 / x.
Triangles QBC and QAD are similar, which means QC = 5x.
QE ^ 2 = QD * QC = 30x ^ 2.
QE = x√30.
Let’s find the distance from point E to line CD:
EP = QE * cos of the angle QEP = Q * cos of the angle QDA = QE * cos α = x√30 * 1 / x = √30.
Problem answer: EP = √30.