2014?咸宁)如图,在△ABC中,AB=AC=10,点D是边BC上一动点(不与B,C重合),
①∵AB=AC,∴∠B=∠C,又∵∠ADE=∠B∴∠ADE=∠C,∴△ADE∽△ACD;故①正确,②AB=AC=10,∠ADE=∠B=α,cosα=4 5 ,∴BC=2ABcosB=2×10×4 5 =16,∵BD=6,∴DC=10,∴AB=DC,在△ABD与△DCE中,∠BAD=∠CDE ∠B=∠C AB=DC ∴△ABD≌△DCE(ASA).故②正确,③当∠AED=90°时,由①可知:△ADE∽△ACD,∴∠ADC=∠AED,∵∠AED=90°,∴∠ADC=90°,即AD⊥BC,∵AB=AC,∴BD=CD,∴∠ADE=∠B=α且cosα=4 ...全部
①∵AB=AC,∴∠B=∠C,又∵∠ADE=∠B∴∠ADE=∠C,∴△ADE∽△ACD;故①正确,②AB=AC=10,∠ADE=∠B=α,cosα=4 5 ,∴BC=2ABcosB=2×10×4 5 =16,∵BD=6,∴DC=10,∴AB=DC,在△ABD与△DCE中,∠BAD=∠CDE ∠B=∠C AB=DC ∴△ABD≌△DCE(ASA).故②正确,③当∠AED=90°时,由①可知:△ADE∽△ACD,∴∠ADC=∠AED,∵∠AED=90°,∴∠ADC=90°,即AD⊥BC,∵AB=AC,∴BD=CD,∴∠ADE=∠B=α且cosα=4 5 ,AB=10,BD=8.当∠CDE=90°时,易△CDE∽△BAD,∵∠CDE=90°,∴∠BAD=90°,∵∠B=α且cosα=4 5 .AB=10,∴cosB=AB BD =4 5 ,∴BD=25 2 .故③正确.④易证得△CDE∽△BAD,由②可知BC=16,设BD=y,CE=x,∴AB DC =BD CE ,∴10 16?y =y x ,整理得:y2-16y 64=64-10x,即(y-8)2=64-10x,∴0<x≤6。
4.故④正确.故答案为:①②③④。收起