Identifying a Kite: Comparing Quadrilaterals with Equal Sides and Equilateral Triangles

Let's organize the data systematically according to the order specified in the problem:

a. $AB=AD$b.$$$GE\perp HF$c. $$$$$DB=BC=CD$d. $EF=FH=HE$Let's continue and refer to square ABCD. From facts a and c mentioned above, it's clearly evident that this is a square with two (different) pairs of equal adjacent sides:

$$

And this is the definition of a kite, therefore - square ABCD is indeed a kite.

Let's continue and refer to square EFGH and note two important facts:

The fact that triangle EFH is equilateral (fact d) and the fact that the diagonals of square EFGH are perpendicular to each other (fact b),

Let's therefore draw square EFGH separately and extend its diagonals EG and HF (meaning - connect points E and G and points H and F), let's mark the intersection point of the diagonals with the letter M:

Note that segment EM is a height to side HF in the equilateral triangle EFH, and therefore also a median to that side in triangle EFH (this follows from the theorem stating that the height, median, and angle bisector to the same side in an equilateral triangle coincide), meaning:

e.

$HM=MF$

Therefore we found that in square EFGH the main diagonal (EG) bisects the secondary diagonal (HF) and is perpendicular to it,

Let's continue and refer to the fact that we can also write the given about the perpendicular diagonals in square EFGH as follows:

f. $MG\perp HF$and therefore in triangle HGF there is also a median to the side (from e) which is also a height to it (from f),

And from the theorem: "If in a triangle the median to a side is also a height to that side - the triangle is isosceles" (and the side is the base of the isosceles triangle) we conclude:

g.

$HG=FG$Therefore from facts g and d we conclude that square EFGH is indeed a kite (since it meets the definition of a kite mentioned earlier)

Therefore the correct answer is answer a.