Giải bài 53 trang 89 – SGK Toán lớp 9 tập 2

Hướng dẫn:

Sử dụng định lý: "Trong một tứ giác nội tiếp, tổng số đo hai góc đối nhau bằng $180^o$"

TH1: $\left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & {{80}^{o}}+\widehat{C}={{180}^{o}} \\ & {{70}^{o}}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{C}={{100}^{o}} \\ & \widehat{D}={{110}^{o}} \\ \end{aligned} \right.$

TH2: $\left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{A}+{{105}^{o}}={{180}^{o}} \\ & \widehat{B}+{{75}^{o}}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{A}={{75}^{o}} \\ & \widehat{B}={{105}^{o}} \\ \end{aligned} \right.$

TH3: $\left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & 60^o+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{C}=120^o \\ & \widehat{B}=\alpha \\ &\widehat{D}=180^o-\alpha \end{aligned} \right.$

TH4: $\left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ &40^o+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{A}=\alpha \\ & \widehat{C}=180^o-\alpha \\ &\widehat{D}=140^o \end{aligned} \right.$

TH5: $\left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{A}+{{74}^{o}}={{180}^{o}} \\ & \widehat{D}+{{65}^{o}}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{A}={{106}^{o}} \\ & \widehat{D}={{115}^{o}} \\ \end{aligned} \right.$

TH6: $\left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & {{95}^{o}}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+{{98}^{o}}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{C}={{85}^{o}} \\ & \widehat{B}={{82}^{o}} \\ \end{aligned} \right.$

Từ đó, ta có kết quả trong bảng sau: