(1)连CD
∵△ABC是等腰直角△,∴CD⊥AB,且AD=CD=BD
又AE=CF,∠EAD=∠FCD,边角边,△FDC全等于△EAD
所以∠EDA=∠FDC ∠CDA=∠CDE+∠EDA=∠CDE+∠FDC=∠FDE=90°
又△FCD全等于△EAD,ED=FD
△DEF是等腰直角三角形
(2)不会变.
∵△FCD全等于△EAD,S△FCD=S△EAD
S四边形CEDF=S△FCD+S△CDE=S△EAD+S△CDE=S△CAD=1/2S△ABC
(1)连CD
∵△ABC是等腰直角△,∴CD⊥AB,且AD=CD=BD
又AE=CF,∠EAD=∠FCD,边角边,△FDC全等于△EAD
所以∠EDA=∠FDC ∠CDA=∠CDE+∠EDA=∠CDE+∠FDC=∠FDE=90°
又△FCD全等于△EAD,ED=FD
△DEF是等腰直角三角形
(2)不会变.
∵△FCD全等于△EAD,S△FCD=S△EAD
S四边形CEDF=S△FCD+S△CDE=S△EAD+S△CDE=S△CAD=1/2S△ABC