1、由已知可得:∠BAC=∠BAC'=∠B'AC=150.
∠B'AC‘=∠BAC+∠BAC'+∠B'AC-360
=150×3-360
=90.
所以C'A垂直于B'A.
2、因为∠C'Fc是三角形BFC的外角.所以∠C'FC=∠C'BC+∠FCB
=2∠ABC+2∠ACB
=2(∠ABC+∠ACB)
=2(180-∠ABC)
=2(180-150)
=60.
1、由已知可得:∠BAC=∠BAC'=∠B'AC=150.
∠B'AC‘=∠BAC+∠BAC'+∠B'AC-360
=150×3-360
=90.
所以C'A垂直于B'A.
2、因为∠C'Fc是三角形BFC的外角.所以∠C'FC=∠C'BC+∠FCB
=2∠ABC+2∠ACB
=2(∠ABC+∠ACB)
=2(180-∠ABC)
=2(180-150)
=60.