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duong612009 · Answer

Đáp án`+`Giải thích các bước giải:
Xét `	riangleABE` có:
`CB = CA = CE` (gt)
`=> 	riangleABE` vuông tại `B` 
Xét `	riangleACB` có:
`AC = CB = AB`
`=> 	riangleACB` đều
`=> \hat{CBA} = 60^o`
Ta có: `\hat{ABE} = 90^o` (do `	riangleABE` vuông tại `B`)
`=> \hat{CBA} + \hat{CBE} =90^o`
`=> 60^o +x=  90^o`
`=> x= 30^o`
Vậy `x =3 0^o`
`@`duong612009

Longhack1805 · Answer

Đáp án`+`Giải thích các bước giải :
`triangle ABE` có `BC=AC=EC=1/2AE`
`=>triangleABE` vuông tại `B`
`=>hat{A}+hat{E}=90^o`
`=>hat{E}=90-hat{A}`
`=>hat{E}=90-60^o=30^o`
Mà `triangle BCE` có `BC=EC`
`=>triangleBCE` cân tại `C`
`=>hat{E}=hat{CBE}=30^o`
Hay `x=30^o`

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