Bạn tham khảo!
Qua E kẻ tia Ea sao Ea//Ax
$→$ ∠A + ∠E1 = $180^{o}$ (2 góc tcp)
$→$ ∠E1 = $180^{o}$ - $140^{o}$
$→$ ∠E1 = $40^{o}$
Vì Ax//Cy; Ea//Ax $→$ Cy//Ea (quan hệ từ ⊥ đến //)
$→$ ∠E2 + ∠C = $180^{o}$ (2 góc tcp)
$→$ ∠E2 = $180^{o}$ - $110^{o}$
$→$ ∠E2 = $70^{o}$
Ta có : ∠AEC = ∠E1 + ∠E2 = $40^{o}$ $+$ $70^{o}$ = $110^{o}$
$FbBinhne2k88$