Question

如图所示，三角形ABC中，AB=AC，AD=BD，AC=DC，求角B

Answer 1

∵AB=AC，∴ ∠B=∠C，
∵AD=BD，∴∠BAD=∠B，
∵AC=CD，
∴∠CAD=1/2(180°-∠C)=90°-1/2∠B，
三角形内角和为180°得：
(∠B+90°-1/2∠B)+∠B+∠B=180°，
2.5∠B=90°，
∠B=36°。

Answer 2

∵AB=AC,AD=BD,AC=CD 所以∠B=∠C,∠B=∠BAD;∠CAD=∠C
∴4∠B=180
∠B=45°

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Answer 3

∵AB=AC，AD=BD，AC=DC
∴∠B=∠C=∠BAD=∠CAD
∵∠BAC+∠B+∠C=180°
∴∠BAD+∠CAD+∠B+∠C=180°
4∠B=180°
∠B=45°

Answer 4

∵∴∠

∵ AB = AC
∴ ∠C = ∠B
∵ DB = AB
∴ ∠BAD = ∠B
∵ AC = AD
∴ ∠DAC = ∠ADC = ∠B + ∠BAD = 2∠B

∵ ∠BAC + ∠B + ∠C = 180°

∴ ∠BAD + ∠CAD + ∠B + ∠C = ∠B + 2∠B + ∠B + ∠B = 5∠B = 180°

∴ ∠B = 36°