In the figure, ABE is a straight line. The angle bisectors of ∠BAC and ∠CBE intersect at D. AD and BC intersect at F. Prove that ∠ADB = 1/2∠ACB
圖片參考：http://imgcld.yimg.com/8/n/HA00386248/o/701101010028213873434910.jpg
Prove triangle (angle)
2011-01-01 7:55 pm
回答 (3)
2011-01-01 8:23 pm
✔ 最佳答案
Let α=CAD=BAD, β =CBD=EBD.Consider ΔABD,
∟ABD+∟BAD+∟ADB=180 (∟ sum of triangle)
(180-β)+α+∟ADB=180
∟ADB=β-α
Consider ΔABC,
∟ABC+∟BAC+∟ACB=180(∟sum of triangle)
(180-2β)+2α+∟ACB=180
∟ACB=2(β-α)
So ∟ADB=1/2 ∟ACB
2011-01-05 9:57 pm
點解你唔答既????
2011-01-01 9:59 pm
BAF=x,DBF=y:
x+ADB=y(ext. angle of triangle)
ADB=y-x
________________
ACB+2x=2y(ext. angle of triangle)
ACB=(2y-2x)
ACB=2(y-x)
_____________________
ADB/ACB=(y-x)/2(y-x)
ADB/ACB=1/2
ADB=(ACB/2)
x+ADB=y(ext. angle of triangle)
ADB=y-x
________________
ACB+2x=2y(ext. angle of triangle)
ACB=(2y-2x)
ACB=2(y-x)
_____________________
ADB/ACB=(y-x)/2(y-x)
ADB/ACB=1/2
ADB=(ACB/2)
收錄日期: 2021-04-19 23:49:07
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