## Elementary Geometry for College Students (7th Edition) Clone

$EB=6$
We are given that $AD=5$ and $DE=4$. Altitude $DE$ forms a right triangle, $\triangle ADE$. Using the Pythagorean Theorem... $a^{2}+b^{2}=c^{2}$ $a^{2}+4^{2}=5^{2}$ $a^{2}+16=25$ Subtract 16 from both sides... $a^{2}=9$ Square root both sides... $a=3$ or $AE=3$ Opposite sides of a parallelogram are congruent. Therefore, $DC=AB=9$. Using the Segment Addition Postulate... $AE+EB=AB$ Substitute in known lengths... $3+EB=9$ Subtract three from both sides... $EB=6$