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match each sentence with the correct set - up. not all set - ups will b…

Question

match each sentence with the correct set - up. not all set - ups will be used.
if ∠a and ∠b are vertical angles, then ____.
if d is in the interior of ∠abc, then ____.
if m is the midpoint of (overline{ab}), then ____.
if ∠a and ∠b form a linear pair, then ____.
if (overline{ad}) bisects ∠bac, then ____.
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(mangle a + mangle b = 180^circ), (mangle a + mangle b = 90^circ), (angle a cong angle b), (mangle a = mangle b), (angle bad cong angle dac), (mangle bad = mangle dac), (mangle abd + mangle dbc = mangle abc), (overline{am} cong overline{mb}), (am = mb), (overline{ab} cong overline{bc}), (ab = bc), (mangle abc + mangle bcd = mangle abd)

Explanation:

Step1: Analyze "If ∠A and ∠B are vertical angles"

Vertical angles are equal, so \( \angle A \cong \angle B \), \( m\angle A = m\angle B \).

Step2: Analyze "If D is in the interior of ∠ABC"

By angle addition postulate, \( m\angle ABD + m\angle DBC = m\angle ABC \).

Step3: Analyze "If M is the midpoint of \( \overline{AB} \)"

Midpoint divides a segment into two congruent segments, so \( \overline{AM} \cong \overline{MB} \), \( AM = MB \).

Step4: Analyze "If ∠A and ∠B form a linear pair"

Linear pair angles sum to \( 180^\circ \), so \( m\angle A + m\angle B = 180^\circ \).

Step5: Analyze "If \( \overline{AD} \) bisects \( \angle BAC \)"

Angle bisector divides an angle into two congruent angles, so \( \angle BAD \cong \angle DAC \), \( m\angle BAD = m\angle DAC \).

Answer:

  • If \( \angle A \) and \( \angle B \) are vertical angles, then \( \boldsymbol{\angle A \cong \angle B, m\angle A = m\angle B} \).
  • If \( D \) is in the interior of \( \angle ABC \), then \( \boldsymbol{m\angle ABD + m\angle DBC = m\angle ABC} \).
  • If \( M \) is the midpoint of \( \overline{AB} \), then \( \boldsymbol{\overline{AM} \cong \overline{MB}, AM = MB} \).
  • If \( \angle A \) and \( \angle B \) form a linear pair, then \( \boldsymbol{m\angle A + m\angle B = 180^\circ} \).
  • If \( \overline{AD} \) bisects \( \angle BAC \), then \( \boldsymbol{\angle BAD \cong \angle DAC, m\angle BAD = m\angle DAC} \).