如图，在△ABC中，AD平分∠BAC，AD的垂直平分线交AD于E，交BC的延长线于F，角B=40°，求角CAF的度数

---

∠BAD+∠B=∠ADF Exterior angle theorem （外角定岁雹理）乎芦帆

Because EF is AD's Perpendicular bisector（垂直平分线）

So AF=DF
So ∠ADF=∠DAF
∠DAF=∠DAC+∠CAF
And
∠BAD=∠DAC

∠DAC+∠CAF=∠DAF
∠BAD+∠B=∠ADF
So
∠CAF=∠哗粗B=40°

---

是40°

---

40

---