The bases BC and AD of trapezoid ABCD are 4 and 64, respectively, BD = 16, Proof that triangle CBD and BDA are similar.

Let’s formally write the data from the task:

BC = 4, AD = 64, BD = 16.

Note that the angles CBD and BDA are equal, because straight lines AD and BC are parallel:

CBD = BDA = a.

Consider a triangle ABD. By the cosine theorem, we have that

AB ^ 2 = BD ^ 2 + AD ^ 2 – 2 * BD * AD * cos (a) =

= 16 ^ 2 + 64 ^ 2 – 2 * 16 * 64 * cos (a).

Similarly, from triangle BCD, we obtain

CD ^ 2 = BC ^ 2 + BD ^ 2 – 2 * BC * BD * cos (a) =

= 4 ^ 2 + 16 ^ 2 – 2 * 4 * 16 * cos (a).

Then

AB ^ 2 / CD ^ 2 = (16 ^ 2 + 64 ^ 2 – 2 * 16 * 64 * cos (a)) / (4 ^ 2 + 16 ^ 2 – 2 * 4 * 16 * cos (a)) = 4 ^ 2 and

AB / CD = 4. But BD / BC = AD / BD = 4.

So we got that for triangles CBD and BDA