∵AD是∠BAC的平分线,∴∠BAD=∠DAC=∠BAC/2,
∵EF是AD的垂直平分线,∴DE=AE,
还有∠DAE=∠ADE=∠B+∠BAD=∠B+∠BAC/2,
于是∠CAE=∠DAE-∠DAC=∠B+∠BAC/2-∠BAC/2=∠B,
∵△CAE和△ABE中,已证∠CAE=∠ABE,还有公用角∠E,
∴△CAE∽△ABE,得AE/BE=CE/AE,或AE²= BE*CE,
从而DE²=AE²=BE*CE.