已知△ABC的∠B和∠C的平分线BE,CF交于点G.求证:(1)∠BGC=90°+½∠A
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已知△ABC的∠B和∠C的平分线BE,CF交于点G.求证:(1)∠BGC=90°+½∠A
已知△ABC的∠B和∠C的平分线BE,CF交于点G.求证:(1)∠BGC=90°+½∠A
已知△ABC的∠B和∠C的平分线BE,CF交于点G.求证:(1)∠BGC=90°+½∠A
∵∠B=∠C
∴∠EBC=∠FCB=1/2∠ABC=1/2∠ACB
∠ABC+∠ACB=180°-∠A
∠BGC =180°-∠EBC-∠FCB
=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
(1) ∠FGB=∠GBC+∠GCB
∠GBC=1/2∠ABC ∠GCB=1/2∠ACB
∠BGC=180-∠FGB
=180-(∠GBC+∠GCB)
180-1/2(∠ABC+∠ACB)
(2)∠ABC+∠ACB=180-∠A
由(1)得
∠BGC=180-1/2(∠ABC+∠ACB)
=180-1/2(180-∠A)
=90+1/2∠A
因为∠A +∠B + ∠C=180°,180°=∠GCB+∠GBC+∠BGC,又因BE、CF是角平分线,所以:1/2(∠C+∠B)=1/2(180°-∠A) ,1/2(∠C+∠B)=90°+1/2∠A ,1/2(∠C+∠B)=∠GCB+∠GCB,所以,∠CGB=180°-∠GCB-∠GBC,所以∠BGC=90°+1/2∠A (上面那家明显做错了)
上面那个真的做错了,为什么∠B=∠C