Problem:
Square ABCD has side length 13, and points E and F are exterior to the square such that BE=DF=5 and AE=CF=12. Find EF2.
Solution:
Extend AE past A and DF past D to meet at G. Note that ∠ADG=90∘− ∠CDF=∠DCF=∠BAE and ∠DAG=90∘−∠BAE=∠ABE. Thus △AGD≅△BEA. Therefore EG=FG=17, and because ∠EGF is a right angle, EF2=2⋅172=578​.
The problems on this page are the property of the MAA's American Mathematics Competitions
