The problem asks us to find the perimeter of kite $ABCD$. We are given that $AE = EC = 8$, $BE = 4$, and $DE = 16$. We have already found that $CD = 8\sqrt{5}$.

1. Problem Description

The problem asks us to find the perimeter of kite $ABCD$. We are given that $AE = EC = 8$, $BE = 4$, and $DE = 16$. We have already found that $CD = 8\sqrt{5}$.

2. Solution Steps

The perimeter of kite $ABCD$ is $AB + BC + CD + DA$. Since $AB = BC$ and $AD = CD$, the perimeter is $2(AB + CD)$.

First, we need to find $AB$. In right triangle $ABE$, we have $AE = 8$ and $BE = 4$.

By the Pythagorean theorem, $AB^2 = AE^2 + BE^2$.

$AB^2 = 8^2 + 4^2 = 64 + 16 = 80$.

$AB = \sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5}$.

Next, we already know $CD = 8\sqrt{5}$.

The perimeter is $2(AB + CD) = 2(4\sqrt{5} + 8\sqrt{5}) = 2(12\sqrt{5}) = 24\sqrt{5}$.

3. Final Answer

$24\sqrt{5}$