如图,在三角ABC中,∠A=60度,三角ABC的内角平分线或外角平分线交于点P,求∠P
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如图,在三角ABC中,∠A=60度,三角ABC的内角平分线或外角平分线交于点P,求∠P
图(1)∠P=180°-1/2∠ABC-1/2∠ACB
=180°-1/2(∠ABC+1/2∠ACB)
=180°-1/2(180°-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A
图(2)BC延长至点E,假设AC线和BP线相交,交点为D.
∠A=180°-∠ADB-1/2∠ABC
=180°-∠ADB-1/2(180°-∠A-∠ACB)
=180°-∠ADB-90°+1/2∠A+1/2∠ACB
1/2∠A =90°-∠ADB+1/2∠ACB
∠A =180°-2∠ADB+∠ACB
同样道理
∠P =180°-∠PDC-1/2∠ACE
=180°-∠PDC-1/2(180°-∠ACB)
=180°-∠PDC-90°+1/2∠ACB
=90°-∠PDC+1/2∠ACB
由于∠ADB=∠PDC
所以∠A =180°-2∠PDC+∠ACB
而∠P =90°-∠PDC+1/2∠ACB
所以∠A =2∠P
图(3)
∠P=180°-∠CBP-∠BCP
=180°-1/2(180°-∠ABC)-1/2(180°-∠ACB)
=180°-90°+1/2∠ABC-90°+1/2∠ACB
=1/2∠ABC+1/2∠ACB
=1/2(∠ABC+∠ACB)
=1/2(180°-∠A)
=90°-1/2∠A
=180°-1/2(∠ABC+1/2∠ACB)
=180°-1/2(180°-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A
图(2)BC延长至点E,假设AC线和BP线相交,交点为D.
∠A=180°-∠ADB-1/2∠ABC
=180°-∠ADB-1/2(180°-∠A-∠ACB)
=180°-∠ADB-90°+1/2∠A+1/2∠ACB
1/2∠A =90°-∠ADB+1/2∠ACB
∠A =180°-2∠ADB+∠ACB
同样道理
∠P =180°-∠PDC-1/2∠ACE
=180°-∠PDC-1/2(180°-∠ACB)
=180°-∠PDC-90°+1/2∠ACB
=90°-∠PDC+1/2∠ACB
由于∠ADB=∠PDC
所以∠A =180°-2∠PDC+∠ACB
而∠P =90°-∠PDC+1/2∠ACB
所以∠A =2∠P
图(3)
∠P=180°-∠CBP-∠BCP
=180°-1/2(180°-∠ABC)-1/2(180°-∠ACB)
=180°-90°+1/2∠ABC-90°+1/2∠ACB
=1/2∠ABC+1/2∠ACB
=1/2(∠ABC+∠ACB)
=1/2(180°-∠A)
=90°-1/2∠A
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