(1)∠B=90℃,AB=BC =4,由勾股定理得AC=4√2,
(2)∵AD =2,CD =6,AC=4√2,
∴AD²+CD²=AC²
由勾股定理的逆定理得△ACD是直角三角形.
(3)△ACE面积为4,BC =4,所以BC×AE/2=4,得AE=2.BE=AB-AE=4-2=2
(1)∠B=90℃,AB=BC =4,由勾股定理得AC=4√2,
(2)∵AD =2,CD =6,AC=4√2,
∴AD²+CD²=AC²
由勾股定理的逆定理得△ACD是直角三角形.
(3)△ACE面积为4,BC =4,所以BC×AE/2=4,得AE=2.BE=AB-AE=4-2=2