Kẻ $ BE⊥CD$
Ta có: $ABED$ là hình chữ nhật $\Rightarrow BE=AD=12cm$
Do $ΔBCD$ vuông tại $B$ có $BE⊥DC$ nên ta có:
$\left \{ {{BD^2+BC^2=CD^2} \atop {BD.BC=BE.CD}} \right.$ $\Rightarrow \left \{ {{BD^2+BC^2=625} \atop {BD.BC=300}} \right.$ $\Rightarrow \left \{ {{(BD+BC)^2=1225} \atop {BD.BC=300}} \right.$ $\Rightarrow \left \{ {{BD+BC=35} \atop {BD.BC=300}} \right.$ $\Rightarrow \left[ \begin{array}{l}\left \{ {{BC=15} \atop {BD=20}} \right.\\\left \{ {{BC=20} \atop {BD=15}} \right.\end{array} \right.$
`+)`Nếu $BD=20$ thì $AB=\sqrt{BD^2-AD^2}=\sqrt{20^2-12^2}=16$
`+)`Nếu $BD=15$ thì $AB=\sqrt{BD^2-AD^2}=\sqrt{15^2-12^2}=9$
