问题补充:
如图:在直角梯形ABCD中,AB//CD,AD⊥DC,AB=BC,且AE⊥BC.(1)若AD=8,DC=4,求AB的长
答案:
(1)连接AC,
∵AB∥CD,
∴∠ACD=∠BAC,
∵AB=BC,
∴∠ACB=∠BAC,
∴∠ACD=∠ACB,
∵AD⊥DC,AE⊥BC,
∴∠D=∠AEC=90°,
∵AC=AC,
∴ ,∴△ADC≌△AEC,(AAS)
∴AD=AE;
(2)由(1)知:AD=AE,DC=EC,
设AB=x,则BE=x-4,AE=8,
在Rt△ABE中∠AEB=90°,
由勾股定理得:82+(x-4)2=x2,
解得:x=10,
∴AB=10.
======以下答案可供参考======
供参考答案1:
(1)连接AC,
∵AB∥CD,
∴∠ACD=∠BAC,
∵AB=BC,
∴∠ACB=∠BAC,
∴∠ACD=∠ACB,
∵AD⊥DC,AE⊥BC,
∴∠D=∠AEC=90°,
∵AC=AC,
∴ ,∴△ADC≌△AEC,(AAS)
∴AD=AE
供参考答案2:
(1)连接AC
∵AB∥CD
∴∠ACD=∠BAC
∵AB=BC
∴∠ACB=∠BAC
∴∠ACD=∠ACB
∵AD⊥DC AE⊥BC
∴∠D=∠AEC=900
∵AC=AC
∴△ADC≌△AEC
∴AD=AE
(2)由(1)知:AD=AE ,DC=EC
设AB=x, 则BE=x-4 ,AE=8
在Rt△ABE中 ∠AEB=900由勾股定理得:解得:x=10∴AB=10供参考答案3:(1)证明:连接AC,∵AB∥CD,
∴∠ACD=∠BAC,
∵AB=BC,
∴∠ACB=∠BAC,
∴∠ACD=∠ACB,
∵AD⊥DC,AE⊥BC,
∴∠D=∠AEC=90°,
∵AC=AC,
∴∠D=∠AEC∠DCA=∠ACBAC=AC,∴△ADC≌△AEC,(AAS)
∴AD=AE;
(2)由(1)知:AD=AE,DC=EC,设AB=x,则BE=x-4,AE=8,在Rt△ABE中∠AEB=90°,由勾股定理得:8²+(x-4)²=x²,解得:x=10,∴AB=10.
供参考答案4:(1)连接AC,
∵AB∥CD,
∴∠ACD=∠BAC,
∵AB=BC,
∴∠ACB=∠BAC,
∴∠ACD=∠ACB,
∵AD⊥DC,AE⊥BC,
∴∠D=∠AEC=90°,