Question

三角形ABC是等腰三角形,其中AB=AC,角B的角平分线交AC于D,且BC=BD+AD,求角A

Answer 1

做DE||CB,交AB于E.
在BC上取一点F,使 BF = BD,FC = AD.
ED||BC,
所以,内错角相等,∠DBC = ∠BDE
BD是角平分线,所以 ∠DBC = ∠DBE = ∠BDE
所以△EBD 是等腰三角形,EB = ED
ED||BC,AB=AC,所以
EB = DC
所以 ED = DC
ED||BC,所以同位角相等,∠ADE = ∠C
在△AED 和 △FDC中
AD = FC
∠ADE = ∠C
ED = DC
所以,根据边角边关系,二者全等.
所以
FD =AE = AD = FC
△FDC 也是等腰三角形
∠FDC = ∠C
另外 ∠A = ∠DFC
BD = BF,所以 △BDF 是等腰三角形
∠BDF ∠BFD
根据三角形内角和180,所以
∠ADE = (180 - ∠A)/2
∠EDB = ∠ABD = ∠B/2 = (180-∠A)/4
∠BDE = ∠BFD = 180 - ∠DFC = 180 - ∠A
∠FDC = ∠C = (180 -∠A)/2
以上四个角之和为 180度.所以
(180 -∠A)/2 + (180-∠A)/4 + 180 -∠A + (180-∠A)/2 = 180
(180-∠A)*(1/2 + 1/4 + 1 + 1/2) = 180
(180 - ∠A)*9/4 = 180
180 -∠A = 80
∠A = 100