ABCD is a kite. $AC$ is perpendicular to $DB$, and $DE = EB$. We are given $AE = 10$ cm, $BE = 8$ cm, and $CE = 9$ cm. We need to find the length of $AC$.

1. Problem Description

ABCD is a kite. $AC$ is perpendicular to $DB$, and $DE = EB$. We are given $AE = 10$ cm, $BE = 8$ cm, and $CE = 9$ cm. We need to find the length of $AC$.

2. Solution Steps

Since $AC$ is perpendicular to $DB$, triangle $ABE$ is a right triangle. We have $AE = 10$ cm and $BE = 8$ cm.

Using the Pythagorean theorem in triangle $ABE$, we have

$AB^2 = AE^2 + BE^2 = 10^2 + 8^2 = 100 + 64 = 164$.

So $AB = \sqrt{164} = 2\sqrt{41}$ cm.

Also, triangle $CBE$ is a right triangle. We have $CE = 9$ cm and $BE = 8$ cm.

Using the Pythagorean theorem in triangle $CBE$, we have

$BC^2 = CE^2 + BE^2 = 9^2 + 8^2 = 81 + 64 = 145$.

So $BC = \sqrt{145}$ cm.

In a kite, two pairs of adjacent sides are equal. So $AB = AD$ and $BC = CD$. Thus $AD = \sqrt{164}$ and $CD = \sqrt{145}$.

Consider triangle $ADE$. $AE = 10$, $DE = 8$, $AD = \sqrt{164}$.

Consider triangle $CDE$. $CE = 9$, $DE = 8$, $CD = \sqrt{145}$.

The length of $AC = AE + EC = 10 + 9 = 19$ cm.

3. Final Answer

19 cm