∵AB=AC,BC=BD,AD=DE=EB,
∴∠A=∠AED,∠ABC=∠C=∠BDC,∠EDB=∠EBD
設(shè)∠A=x,因∠AED=∠EDB+∠EBD
故∠EDB=∠EBD=x/2
∵∠BDC=∠A+∠ABD=x+x/2=3x/2
∴∠BDC=∠C=∠ABC=3x/2
∵∠A+∠ABC+∠C=180°
∴X+3X/2*2=180
得X=45°
∴∠A的度數(shù)是45°
這個事道數(shù)學(xué)幾何題 問題;三角形△ABC AB=AC BC=BD AD=DE=EB 求∠A的度數(shù)?
這個事道數(shù)學(xué)幾何題 問題;三角形△ABC AB=AC BC=BD AD=DE=EB 求∠A的度數(shù)?
數(shù)學(xué)人氣:983 ℃時間:2019-11-07 20:02:57
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