Question

几何题.快来啊如图,△ABC的∠B和∠C的平分线BE,CF相交于G,求证：（1）∠BGC=180°-2分之1（∠ABC+

Answer 1

∠BGC=180°-(∠GBC+∠GCB)=180°-(∠ABC/2+∠ACB/2)=180°-(∠ABC+∠ACB)/2

∠BGC=180°-(∠ABC+∠ACB)/2=180°-(180°-∠A)/2=90°+∠A/2