Подібні трикутники та трапеції

Similar triangles in a trapezoid? What?

"But there aren't any triangles in a trapezoid to begin with!" you would ask. And you are absolutely correct:

I don't see any triangles... but what if... we draw some extra lines?

Let's extend $\overline{AD}$ and $\overline{BC}$ so that they meet at a point $E$:

Now do you see the similar triangles?

Since $\overline{AB} || \overline{CD}$ we have $\angle{BAE} \cong \angle{CDE}$ and $\angle{EBA} \cong \angle{ECD}$. By the AA-similarity theorem, we now have $\triangle ABE \sim \triangle DCE$.

This is a really useful result! I'll leave it to you to figure out the rest of the results you can draw from this.

Now, let's draw $\overline{AC}$ and $\overline{BD}$ instead!

You can now easily deduce that $\triangle BAE \sim \triangle DCE$!

These results are extremely useful when dealing with trapezoids. When you have a trapezoid and you have the intersection of its diagonals or the extension of its legs, consider similar triangles!

Here is a practice problem for your new skill! Good luck!