The bases BC and AD of trapezoid ABCD are 7 and 28, respectively. BD = 14, prove that triangles CBD BDA are similar.

We will look for one of three signs of similarity of triangles.

Here is one of the signs of similarity:

two triangles are similar if they have an equal angle, and the sides that form this angle in one triangle are proportional to the other two sides in the second triangle.

Consider in this trapezoid:

<ADB = <DBC, as crosswise lying angles with parallel lines AD and BC, and secant BD.

Next, consider the sides of these corners:

AD / BD = 28/14 = 2, BD / BC = 14/7 = 2.

That is, the sign of similarity is observed, and triangles ADB and DCB are similar.