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name _ quiz 2.1 - 2.4 period _ date _ e is the mid - point of df. find …

Question

name _ quiz 2.1 - 2.4 period _ date _ e is the mid - point of df. find the segment lengths. 1. find de and ef. 2. find de and df. b is the mid - point of ac. 3. find the value of x. $overrightarrow{bd}$ bisects $angle abc$. find the angle measure. 4. find $angle abd$. 5. find $angle dbc$.

Explanation:

Step1: Recall mid - point property

If E is the mid - point of DF and DF = 30, then DE=EF. Since DE + EF=DF and DE = EF, we have DE=EF=\frac{DF}{2}.
DE = EF=\frac{30}{2}=15

Step2: Recall mid - point property for second segment

If E is the mid - point of DF and EF = 7, then DE = EF = 7. And DF=DE + EF, so DF=7 + 7=14

Step3: Use mid - point property for variable

If B is the mid - point of AC, then AB = BC. Given AB = 3x and BC = 27, we set up the equation 3x=27. Solving for x, we divide both sides by 3: x=\frac{27}{3}=9

Step4: Use angle - bisector property

If \(\overrightarrow{BD}\) bisects \(\angle ABC\) and \(\angle DBC = 22^{\circ}\), then \(\angle ABD=\angle DBC\). So \(\angle ABD = 22^{\circ}\)

Step5: Use angle - bisector property

If \(\overrightarrow{BD}\) bisects \(\angle ABC\) and \(\angle ABC=109^{\circ}\), then \(\angle DBC=\frac{\angle ABC}{2}\). So \(\angle DBC=\frac{109^{\circ}}{2}=54.5^{\circ}\)

Answer:

  1. DE = 15, EF = 15
  2. DE = 7, DF = 14
  3. x = 9
  4. \(\angle ABD = 22^{\circ}\)
  5. \(\angle DBC = 54.5^{\circ}\)