接着上面的, ∠BAC=110° ,E/G分别为AB、AC的中点。DE⊥AB,FG⊥AC,求∠DAF的度数
据题意可知AD=BD,AC=FC
∠ BAD=∠ ABD,∠ACF = ∠CAF
∠ ABD+∠ACF =180-∠BAC=180-110=70
∠ BAD+∠CAF=∠ ABD+∠ACF=70
∠DAF=∠BAC-(∠ BAD+∠CAF)=110- 70=60