Question

如图，△ABC的∠B和∠C的平分线BE、CF相交于点G， 求证：∠BGC＝90º＋2/1∠A.

Answer 1

∵∠B和∠C的平分线BE、CF相交于点G
∴∠ABE=∠CBE=1/2∠ABC，∠ACF=∠BCF=1/2∠ACB
∵∠BGC=180°-∠CBE-∠BCF
=180°-1/2∠ABC-1/2∠ACB
=180°-1/2(∠ABC+∠ACB)
=180°-1/2(180°-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A
你他妈的抄袭我的。。。。可耻

Answer 2

∵∠B和∠C的平分线BE、CF相交于点G
∴∠ABE=∠CBE=1/2∠ABC，∠ACF=∠BCF=1/2∠ACB
∴∠BGC=180°-∠CBE-∠BCF
=180°-1/2∠ABC-1/2∠ACB
=180°-1/2(∠ABC+∠ACB)
∵∠ABC+∠ACB =180°-∠A
∴180°-1/2(∠ABC+∠ACB)=180°-1/2(180°-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A

Answer 3

在三角形ABC中
∠A+∠ABC+∠ACB=180
在三角形GBC中
∠BGC+∠GBC+∠GCB=180
又因为BG平分... CG平分...
所以∠ABC=2∠GBC ∠ACB=2∠GCB
所以2∠GBC +2∠GCB+2∠GCB=360
所以∠A+180=2∠BGC
所以
90º＋2/1∠A.=∠BGC