Question

已知:如图,在△ABC中,角ABC和角ACB的角平分线相交于点I。求证：角BIC=90°+1/2角A

Answer 1

∵BI平分∠ABC, ∴∠ABI=∠DBI=1/2∠ABC

∵CI平分∠ACB, ∴∠ACI=∠DCI=1/2∠ACB.

∠BID=∠BAI+∠ABI=∠BAI+1/2∠ABC;

∠CID=∠CAI+∠ACI=∠CAI+1/2∠ACB;

而∠BAI+∠CAI=∠A,

 

∴∠BIC=∠BID+∠CID

=∠BAI+1/2∠ABC+∠CAI+1/2∠ACB

=∠A+1/2∠ABC+1/2∠ACB,

=(1/2∠A+1/2∠A)+1/2∠ABC+1/2∠ACB

=(1/2∠A+1/2∠ABC+1/2∠ACB)+1/2∠A

=1/2(∠A+∠ABC+∠ACB)+1/2∠A.......（因为三角形三内角之和=180°）

=1/2(180°)+1/2∠A

=90°+1/2∠A.