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1. given: o is the mid - point of mn, om = ow. prove: ow = on. 2. given…

Question

  1. given: o is the mid - point of mn, om = ow. prove: ow = on.
  2. given: ab = cd. prove: ac = bd.
  3. given: m∠1 = 90°. prove: m∠2 = 90°.
  4. given: ∠1 and ∠2 are complementary, ∠3 and ∠2 are complementary. prove: m∠1 = m∠3.
  5. given: m∠1 = m∠3. prove: m∠jol = m∠kom.
  6. given: m∠1 = 90°. prove: m∠2+90 = 180.
  7. given: pr≅ln, q is the mid - point of pr, m is the mid - point of ln. prove: pq = lm.
  8. given: ef⊥eg, d is in the interior of ∠feg. prove: ∠fed and ∠deg are complementary.
  9. given: ab≅cd. prove: ac≅bd.
  10. given: ∠1 and ∠2 are supplementary, ∠1≅∠2. prove: ∠1 and ∠2 are right angles.
  11. given: ∠1≅∠2. prove: ∠1 and ∠2 are right angles.
  12. given: ∠1 and ∠2 are complementary. prove: ∠2 and ∠3 are complementary.

Explanation:

Response

1.

Step1: Definition of mid - point

Since O is the mid - point of $\overline{MN}$, by the definition of mid - point, $OM = ON$.

Step2: Given condition

Given $OM = OW$.

Step3: Transitive property

By the transitive property of equality (if $a = b$ and $b = c$, then $a = c$), since $OM = ON$ and $OM = OW$, we have $OW=ON$.

Step1: Segment addition postulate

We know that $AC=AB + BC$ and $BD=CD + BC$.

Step2: Given condition

Given $AB = CD$.

Step3: Substitution

Substitute $AB$ for $CD$ in the equation for $BD$. So $BD=AB + BC$. Since $AC=AB + BC$ and $BD=AB + BC$, by the transitive property of equality, $AC = BD$.

Step1: Vertical angles

$\angle1$ and $\angle2$ are vertical angles.

Step2: Vertical - angle theorem

By the vertical - angle theorem, vertical angles are congruent. Since $m\angle1 = 90^{\circ}$ and $\angle1\cong\angle2$, then $m\angle2=90^{\circ}$.

Answer:

The proof is completed as above.

2.