解?$ ∶ $?連接?$ AC$?
∵?$△ABE≌△BCD$?
∴?$AB=BC$?，?$ AE=BD$?，?$ BE=CD$?，?$ ∠BAE=∠CBD$?
∵?$∠ABE+∠BAE=90°$?
∴?$∠ABE+∠CBD=90°$?
∴?$∠ABC=90°$?
∴?$S_{四邊形ABCD}=S_{△ABD}+S_{△BDC}$?
?$=\frac {1}{2}\ \mathrm {B} D ·A E+\frac {1}{2}\ \mathrm {B} D ·C D$?
?$=\frac {1}{2}\ \mathrm {A} E^2+\frac {1}{2}\ \mathrm {A} E ×B E$?
?$=\frac {1}{2}\ \mathrm {A} E^2+\frac {1}{2}\ \mathrm {B} D ×B E$?
∵?$S_{四邊形 ABCD}=S_{△ABC}+S_{△ADC}$?
?$=\frac {1}{2}\ \mathrm {A} B ×B C+\frac {1}{2}\ \mathrm {C} D ×D E$?
?$=\frac {1}{2}\ \mathrm {A} B ×A B+\frac {1}{2}\ \mathrm {B} E ×D E$?
?$=\frac {1}{2}\ \mathrm {A} B^2+\frac {1}{2}\ \mathrm {B} E ×D E$?
∴?$\frac {1}{2}\ \mathrm {A} E^2+\frac {1}{2}\ \mathrm {B} D ×B E=\frac {1}{2}\ \mathrm {A} B^2+\frac {1}{2}\ \mathrm {B} E ×D E$?
∴?$A B^2=B D ×B E-B E ×D E+A E^2$?
∴?$A B^2=B E ×(B D-D E)+A E^2$?
即?$ A B^2=B E^2+A E^2$?
