AE//BD
=>\(\widehat{ABD}+\widehat{BAE}=180^0\)(hai góc trong cùng phía)
=>\(\widehat{BAE}+90^0=180^0\)
=>\(\widehat{BAE}=180^0-90^0=90^0\)
Ta có: AE//BD
=>\(\widehat{AED}=\widehat{yDE}\)(hai góc so le trong)
=>\(\widehat{yDE}=55^0\)
Ta có: \(\widehat{yDE}+\widehat{BDE}=180^0\)(hai góc kề bù)
=>\(\widehat{BDE}+55^0=180^0\)
=>\(\widehat{BDE}=125^0\)
Có `AE` // `BD`(gt)
`=>`\(\widehat{AED}=\widehat{EDy}=55^0\)(2 góc so le trong)
\(\widehat{ABD}+\widehat{BAE}=180^0\)(2 góc trong cùng phía)
`=>`\(90^0+\widehat{BAE}=180^0\)
`=>`\(\widehat{BAE}=90^0\)
Xét tứ giác `ABDE` có:
\(\widehat{A}+\widehat{B}+\widehat{D}+\widehat{E}=180^0\)
`=>90^0 + 90^0 + 55^0`\(+\widehat{D}=360^0\)
`=>`\(\widehat{D}=125^0\)hay\(\widehat{BDE}=125^0\)
góc C = 40 độ. Tính góc B , góc HAB, góc HAC.
