Question

在三角形ABC中,BE,CF分别是三角形ABC的两条外角平分线,AG垂直BE于点G,AH垂直CF于点H,连接GH,若三角

Answer 1

过点A做EF//BC∠FAC=∠ACB2∠ACH+∠ACB=180, ∠CAH+∠ACH=90∠ACB=2∠CAH ∠FAC=2∠CAH∠CAH=∠FAH AH⊥CF可知点H为CF中点,AC=AF同理可知点G为BE中点,AB=AEGH为梯形EBCF的中位线GH=(BC+EF)/2 ...