中三 三角形相關的定理

設∠ABD=∠DBC=x
∵AB=AC(已知),∴∠ABC=∠ACB(等腰Δ的底角)
∴∠ACB=∠ABD+∠DBC=2x
∴∠DBC+∠BDC+∠DCB=180°(Δ內角和)
x+75°+2x=180° ∴x=35°
∴∠BAC=180°-2(2x) (Δ內角和)
=180°-2(70°)
=40°

∵ AB = AC∴ ∠ABC = ∠ACB (base. ∠s, isos. △) ∠ABD = ∠DCB (given) ∠ABC = 2∠ABD ∠ACB = 2∠ABD ∠A + ∠ABC + ∠ACB = 180° (∠sum of △)∠A + 2∠ABD + 2∠ABD = 180° ∠A = 180°- 4∠ABD ∠ABD + ∠A = 75° (ext. ∠ of △)∠ABD + 180°- 4∠ABD = 75° 180° - 3∠ABD = 75°- 3∠ABD = - 105° ∠ABD = 35°∠A = 180°- 4∠ABD∠A = 180°- 4(35°)∠A = 40° ∠BAC = 40°