Bài 1 :
H1 : Kẻ tia OH//AB ta được :
$\widehat{BAO}$+$\widehat{ABH}$=$180^o$ ( trong cùng phía bù nhau)
⇒$\widehat{ABH}$=$180^o$-$135^o$=$45^o$
⇒$\widehat{COH}$=$120^o$-$45^o$=$75^o$
⇒$\widehat{COH}$+$\widehat{OCD}$=$75^o$+$105^o$=$180^o$(trong cùng phía)
⇒OH//CD
⇒OH//CD//AB
H2:Kẻ OH//AB ta được :
$\widehat{BAO}$=$\widehat{AOH}$=$75^o$(so le trong)
⇒$\widehat{HOC}$=$105^o$-$75^o$=$30^o$
⇒$\widehat{HOC}$=$\widehat{OCD}$=$30^o$(so le trong)
⇒OH//CD//AB
H3: kẻ OH//a
⇒$\widehat{AOH}$=$\widehat{A}$=$37^o$(so le trong)
⇒$\widehat{HOB}$=$62^o$-$37^o$=$25^o$
⇒$\widehat{HOB}$=$\widehat{B}$=$25^o$+$155^o$=$180^o$(trong cùng phía)
⇒OH//b//a
Bài 2 :
H1: Kẻ AH//Cx
$\widehat{xCA}$=$\widehat{CAH}$=$25^o$(so le trong)
⇒$\widehat{BCH}$=$84^o$-$25^o$=$59^o$
Cx//By//AH
⇒$\widehat{ABy}$=$180^o$-$59^o$=$121^o$
H2: Ax//Cy
$\widehat{x}$=$180^o$-$138^o$+$42^o$=$84^o$
H3:
$\widehat{x}$=$180^o$-$140^o$+$35^o$=$75^o$
CHO MK CTLHN NHA:))
