E. B and C are on the same straight line, Ba bisects ∠ EBD, ∠ DBC = 30 ° and calculates the degree of ∠ ABC

E. B and C are on the same straight line, Ba bisects ∠ EBD, ∠ DBC = 30 ° and calculates the degree of ∠ ABC

Because E B C is on the same line and the angle DBC = 30 degrees
So the angle DBE = 180 degrees - 30 degrees = 150 degrees
Because AB bisects the angle EBD, the angle abd = 150 degrees / 2 = 75 degrees
So angle ABC = angle abd + angle DBC = 75 degrees + 30 degrees = 105 degrees

In △ ABC, ∠ BAC = 2 ∠ ACB, D is a point in ⊿ ABC, and ad = CD, BD = ba. Explore the ratio of degree of ∠ DBC to degree of ∠ ABC

Make BM ‖ AC, and make ∠ MCA = ∠ BAC
∵∠ BAC ≠ 90 °∫ MCA + ∠ BAC ≠ 180 °∫ AB is not parallel to cm
And ∵ BM ∥ AC, and ∵ MCA = ∵ BAC ∥ quadrilateral McAb is isosceles trapezoid ∥ AB = cm
∵AD=CD ∴∠DCA=∠DAC
In other words ∵ MCA = ∠ BAC ∵ MCA - ∠ DCA = ∠ BAC - ∠ DAC, that is ∵ MCD = ∠ bad
∴△MCD≌△BAD ∴MD=BD
And ∵ BD = Ba, Ba = MC ∵ MD = BD = Ba = MC
∵∠MCA=∠BAC,∠BAC=2∠ACB ∴∠MCA=2∠ACB ∴∠MCB=∠ACB
∵BM‖AC ∴∠ACB=∠MBC ∴∠MCB=∠MBC ∴MC=MB ∴MB=MD=BD
It is an equilateral triangle. MBD = 60. BCA = MBC = MBD - CBD = 60
∵∠BAC=2∠ACB=2（60°－∠CBD）=120°－2∠CBD
∵ in △ ABC, ∵ BCA + ∵ BAC + ∵ ABC = 180 °
∴（60°－∠CBD）+（120°－2∠CBD）+（∠CBD+∠ABD）=180°∴∠ABD=2∠CBD
∴∠DBC：∠ABC=∠DBC：（∠ABD+∠DBC）=∠DBC：3∠DBC=1：3