Hình tự vẽ.
a, Ta có:
\(\widehat{xAy}=\widehat{x'Ay'}\left(d.d\right)\Rightarrow\widehat{x'Ay'}=36^o\)
\(\widehat{xAy}+\widehat{xAx'}=180^o\Rightarrow\widehat{xAx'}=180^o-36^o=144^o\)
\(\widehat{xAx'}=\widehat{yAy'}\left(d.d\right)\Rightarrow\widehat{yAy'}=144^o\)
b, Ta có:
\(\widehat{xOt}=\widehat{x'Ot'};\widehat{yOt}=\widehat{y'Ot'}\)
mà \(\widehat{xOt}=\widehat{yOt}\left(gt\right)\)
nên \(\widehat{x'Ot'}=\widehat{y'Ot'}\)
=> Ot' là phân giác của \(\widehat{x'Oy'}\).(đpcm)
Chúc bạn hcọ tốt!!!