Bài 56 trang 89

Lời giải

- Tứ giác ABCD là tứ giác nội tiếp => $\widehat{B}+\widehat{D}=180^{\circ}$

$\widehat{ABC}$ = $\widehat{E}$ + $\widehat{ECB}$ (góc ngoài $\Delta EBC$)

                         = $40^{\circ}$ + $\widehat{ECB}$   (1)

$\widehat{ADC}$ = $\widehat{F}$ + $\widehat{DCF}$ (góc ngoài $\Delta FCD$)

                         = $20^{\circ}$ + $\widehat{DCF}$   (2)

=> $\widehat{ABC}+\widehat{ADC}$ = $40^{\circ}+20^{\circ}+\widehat{ECB}+\widehat{DCF}$

=> $180^{\circ}$ = $60^{\circ}+2.\widehat{ECB}$  (Vì $\widehat{ECB}=\widehat{DCF}$)

=> $\widehat{ECB}$ = $60^{\circ}$

Thay vào (1) có: $\widehat{ABC}=40^{\circ}+60^{\circ}=100^{\circ}$

Thay vào (2) có: $\widehat{ADC}=20^{\circ}+60^{\circ}=80^{\circ}$

- $\widehat{BCD}+\widehat{EBC}=180^{\circ}$ (2 góc kề bù)

=> $\widehat{BCD}=180^{\circ}-\widehat{EBC}=180^{\circ}-60^{\circ}=120^{\circ}$

- Tứ giác ABCD là tứ giác nội tiếp => $\widehat{A}\widehat{BCD}=180^{\circ}$

 => $\widehat{A}=180^{\circ}-\widehat{BCD}=180^{\circ}-120^{\circ}=60^{\circ}$