Question

如图,已知三角形abc的两个外角平行线相交于d,求证角bdc=90度-二分之一角a.

Answer 1

因为∠DBC=1/2(180°-∠B)=90°-1/2∠B

∠DCB=1/2(180°-∠C)=90°-1/2∠C

∠DCB+∠DBC+∠BDC=180°

所以∠BDC=180°-（90°-1/2∠B）-（90°-1/2∠C）

=1/2∠B+1/2∠C

=1/2(∠B+∠C)

=1/2(180°-∠A)

=90°-1/2∠A