Theo hình vẽ:

Áp dụng định lý tổng số đo trong tam giác xét tam giác `\DeltaYTX,\DeltaYZT,\DeltaXTK`

`\hat{XYT}+\hat{YTX}+\hat{TXY}=180^o`

`\hat{TYZ}+\hat{YZT}+\hat{YTZ}=180^o`

`\hat{KXT}+\hat{XTK}+\hat{XKT}=180^o`

Suy ra: `\hat{XYT}+\hat{YTX}+\hat{TXY}+\hat{TYZ}+\hat{YZT}+\hat{YTZ}+\hat{KXT}+\hat{XTK}+\hat{XKT}=180^o +180^o +180^o=540^o`

`->\hat{X}+\hat{Y}+\hat{Z}+\hat{T}+\hat{K}=540^o`

Xét tam giác `\DeltaAKD,\DeltaEXC,\DeltaDYB,\DeltaCZA,\DeltaBTE` có:

`\hat{A}+\hat{K}+\hat{D}=180^o`

`\hat{E}+\hat{X}+\hat{C}=180^o`

`\hat{D}+\hat{Y}+\hat{B}=180^o`

`\hat{C}+\hat{Z}+\hat{A}=180^o`

`\hat{B}+\hat{T}+\hat{E}=180^o`

Cộng theo vế tất cả ta được:

`2(\hat{A}+\hat{B}+\hat{C}+\hat{D}+\hat{E})+(\hat{X}+\hat{Y}+\hat{Z}+\hat{T}+\hat{K}=5*180^o`

`->2(\hat{A}+\hat{B}+\hat{C}+\hat{D}+\hat{E})+540^o=900^o`

`->2(\hat{A}+\hat{B}+\hat{C}+\hat{D}+\hat{E})=900^o-540^o=360`

`->\hat{A}+\hat{B}+\hat{C}+\hat{D}+\hat{E}=360^o/2=180^o`

Vậy: ...