Bài 9.
a)

Xét tam giác ABC, ta có:
$\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180{}^\circ$
Mà $\widehat{ABC}=60{}^\circ$; $\widehat{ACB}=25{}^\circ$
$\Rightarrow \widehat{BAC}=95{}^\circ$
$\Rightarrow x+y=95{}^\circ$
Xét tam giác ABD, ta có:
$\widehat{ABD}+\widehat{ADB}+\widehat{DAB}=180{}^\circ$
$\Rightarrow 60{}^\circ +90{}^\circ +y=180{}^\circ$
$\Rightarrow y=30{}^\circ$
$\Rightarrow x=65{}^\circ$
Ta lại có: $\widehat{BAD}=\widehat{ADE}$ $\Rightarrow$ AB // DE (dấu hiệu nhận biết 2 đường thẳng song song)
$\Rightarrow z=\widehat{BAC}$ (2 góc ở vị trí đồng vị do AB // DE)
$\Rightarrow z=95{}^\circ$
Bài 10. Xét tam giác ABC, ta có:
$\widehat{A}+\widehat{B}+\widehat{C}=180{}^\circ$
\[\Rightarrow \widehat{A}+\widehat{B}+\widehat{B}-\widehat{B}+\widehat{C}=180{}^\circ\]
\[\Rightarrow \widehat{A}+2\widehat{B}-(\widehat{B}-\widehat{C})=180{}^\circ\]
\[\Rightarrow \widehat{C}-\widehat{B}=80{}^\circ\]
Bài 11.
a) Xét tam giác ABC, ta có:
$\frac{\widehat{A}}{1}=\frac{\widehat{B}}{3}=\frac{\widehat{C}}{5}=\frac{\widehat{A}+\widehat{B}+\widehat{C}}{9}=\frac{180{}^\circ }{9}=20{}^\circ$
Vậy $\widehat{A}=20{}^\circ$; $\widehat{B}=60{}^\circ$; $\widehat{C}=100{}^\circ$
b)


Ta có:
$\widehat{DBC}=\frac{180{}^\circ -60{}^\circ }{2}=60{}^\circ$
$\widehat{DCB}=180{}^\circ -100{}^\circ =80{}^\circ$
Xét tam giác DBC, dễ dàng CM được: $\widehat{CDB}=40{}^\circ$