.Surely you’ve discovered that the figure below is clipped from a pentagon. Right?
Yeh! let us focus within the border drawn with a crayon, that is, within triangle DAB. How many isosceles are there within it? (For the moment, ignore the smaller pentagram A’C’E’B’D’ and all the triangle formed by their intersections.)
Let’s count it! Recognizing that DA = DB (why?), we will know that DAB, DA’B’ are two isosceles. D’AB is the 3rd isosceles. When connecting points C’ and E’, then DC’E’ and D’C’E’ are also isosceles (having found 5 isos).
There are more though. Through the properties of pentagons, we know that AE’ = BC’ = AB. Therefore ABE’ (with AB = AE’), BAC’ (with BA = BC’) are isosceles. Note that AC’ = C’E’ = E’B, we know that AC’E’ and (with AC’ = C’E’) and C’E’B (with C’E’ = E’B) are isosceles. Finally, AC’ = AD’ and BD’ = BE’ so we find two other isosceles: triangles AC’D’ and BD’E’. Up till now, we have count 11 isosceles in total!
Another nice feature is that, if the original ABCDE is a regular pentagon, then the smaller pentagon A’B’C’D’E’ (as formed by intersection points of the diagonals of the pentagon ABCDE), must also be a regular pentagon. If we look into some of these points A’, B’, C’, D’, E’, we got to find many many more isosceles. But currently let’s ignore them.
We also find a trapezoid A’B’AB. By connecting points C’ and E’, there are other two trapezoids, ABE’C’ and A’B’C’E’. Yes! A’C’AE’ is a trapezoid, it is not an isosceles trapezoid, yet! the shorter base A’C’ equals one of the sides C’A, and is also equal to one diagonal C’E’.
For parallelograms? Either we allow to include some points outside the triangle DAB, or we make more connections (e.g. include the smaller pentagram A’B’C’D’E’); if we do one of the two, then there will be lots of them!