Question

如图，点D在AC上，点E在AB上，且AB=AC,BC=BD，AD=DE=BE，求∠A

Answer 1

AB = AC
∴设∠B = ∠C =x
∵BD = BC
∴∠BDC = ∠C = x
∴∠DBC = 180°-2x
∴∠ABC = ∠DBC - ∠DBC = x-（180°-2x）=3x-180°
∵EB = ED
∴∠EBD = ∠EDB = 3x-180°
∴∠AED = ∠EBD + ∠EDB = 6x-360°
∵AD = ED
∴∠A = ∠AED= 6x-360°
∵∠A+∠ABC+∠C = 180°
∴2x+6x-360°=180°
∴x = 67.5°
∴∠A = 45°

Answer 2

解：设∠EBD=x
∵DE=BE
∴∠AED=2x
又∵AD=DE
∴∠A=2x
∴∠BDC=x+2x=3x
而BC=BD，则∠C=3x
∵AB=AC
∴∠ABC=3x
∴3x+3x+2x=180°
∴∠A=2x=45°．
故填45°．

Answer 3

设∠A=a ∠c=b
因为 AB=AC,所以 ∠ABC=b
在三角形ABC中 a+2b=180 (1);
因为 AD=DE 所以 ∠A=∠AED=a;
又因为DE=BE；所以∠ABD=∠EDB=0.5a
∠ADE+∠EDB+∠BDC=180;即是 180-2a+0.5a+b=180 (2);
由（1）（2）可得a=45