Kẻ CD//AB
Vì CD//AB(ta vẽ)
⇒\(\widehat{A}+\widehat{ACD}=180^O\)(2 góc TCP)
⇒90\(^o\)+\(\widehat{ACD}\) =180\(^o\)
\(\widehat{ACD}=180^o-90^o\)
\(\widehat{ACD}=90^o\)
Ta có: \(\widehat{ACD}+\widehat{DCE}=150^o\)
Thay số:\(90^o+\widehat{DCE}=150^o\)
\(\widehat{DCE}=150^o-90^o\)
\(\widehat{DCE}=60^o\)
Có:\(\widehat{DCE}=60^o\left(cmt\right)\) \(^{\left(1\right)}\)
\(\widehat{CEF}=120^o\left(gt\right)\) \(^{\left(2\right)}\)
Từ \(^{\left(1\right)}\) và\(^{\left(2\right)}\) suy ra:CD//EF( dhnb 2 đường thẳng song song)
Lại có:
AB//CD(ta vẽ)
CD//EF(cmt)
(Từ cả 2 cái trên)⇒AB//EF(đpcm)
Kẻ CD//AB
Vì CD//AB(ta vẽ)
⇒ˆA+ˆACD=180OA^+ACD^=180O(2 góc TCP)
⇒90oo+ˆACDACD^ =180oo
ˆACD=180o−90oACD^=180o−90o
ˆACD=90oACD^=90o
Ta có: ˆACD+ˆDCE=150oACD^+DCE^=150o
Thay số:90o+ˆDCE=150o90o+DCE^=150o
ˆDCE=150o−90oDCE^=150o−90o
ˆDCE=60oDCE^=60o
Có:ˆDCE=60o(cmt)DCE^=60o(cmt) (1)(1)
ˆCEF=120o(gt)CEF^=120o(gt) (2)(2)
Từ (1)(1) và(2)(2) suy ra:CD//EF( dhnb 2 đường thẳng song song)
Lại có:
AB//CD(ta vẽ)
CD//EF(cmt)
(Từ cả 2 cái trên)⇒AB//EF(đpcm).
