如图所示,△ABC中,AD平分∠BAC,EF⊥AD交AB于E,交AC于F,交BC的延长线于H.求证:∠H=0.5(∠AC
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如图所示,△ABC中,AD平分∠BAC,EF⊥AD交AB于E,交AC于F,交BC的延长线于H.求证:∠H=0.5(∠ACB-∠B)
![](http://img.wesiedu.com/upload/d/99/d9977dd16a89e9e99bd1a7ac96d41b86.jpg)
![](http://img.wesiedu.com/upload/d/99/d9977dd16a89e9e99bd1a7ac96d41b86.jpg)
![如图所示,△ABC中,AD平分∠BAC,EF⊥AD交AB于E,交AC于F,交BC的延长线于H.求证:∠H=0.5(∠AC](/uploads/image/z/16186926-30-6.jpg?t=%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA%2C%E2%96%B3ABC%E4%B8%AD%2CAD%E5%B9%B3%E5%88%86%E2%88%A0BAC%2CEF%E2%8A%A5AD%E4%BA%A4AB%E4%BA%8EE%2C%E4%BA%A4AC%E4%BA%8EF%2C%E4%BA%A4BC%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%8EH.%E6%B1%82%E8%AF%81%EF%BC%9A%E2%88%A0H%3D0.5%EF%BC%88%E2%88%A0AC)
∵∠ACB+∠B+∠BAC=180°
∴∠BAC=180°-∠ACB-∠B
∵AD平分∠BAC
∴∠DAC=∠BAC/2
=(180°-∠ACB-∠B)/2
=90°-∠ACB/2-∠B/2
∴∠ADC=180°-(90°-∠ACB/2-∠B/2)-∠ACB
=180°-90+∠ACB/2+∠B/2-∠ACB
=90°-∠ACB/2+∠B/2
∵EF⊥AD
∴∠ADC=90°-∠H
∴∠H=90°-(90°-∠ACB/2+∠B/2)
=90°-90°+∠ACB/2-∠B/2
=0.5(∠ACB-∠B)
∴∠BAC=180°-∠ACB-∠B
∵AD平分∠BAC
∴∠DAC=∠BAC/2
=(180°-∠ACB-∠B)/2
=90°-∠ACB/2-∠B/2
∴∠ADC=180°-(90°-∠ACB/2-∠B/2)-∠ACB
=180°-90+∠ACB/2+∠B/2-∠ACB
=90°-∠ACB/2+∠B/2
∵EF⊥AD
∴∠ADC=90°-∠H
∴∠H=90°-(90°-∠ACB/2+∠B/2)
=90°-90°+∠ACB/2-∠B/2
=0.5(∠ACB-∠B)
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