Question

select the correct answer from each drop - down menu. given: ∠boc and ∠cod are complementary angles; bo intersects ad at point o. prove: ∠aob≅∠bod. statements: 1. ∠boc and ∠cod are complementary angles; bo intersects ad at point o. 2. m∠boc + m∠cod = 90°. 3. m∠bod = 90°. 4. ∠aob and ∠bod are supplementary angles. 5. m∠aob + m∠bod = 180°. 6. m∠aob + 90° = 180°. reasons: 1. given. 2. definition of complementary angles. 3. substitution property of equality. 4. linear pair theorem. 5. definition of supplementary angles. 6. substitution property of equality.

Explanation:

Step1: Recall complementary - angles definition

Since $\angle BOC$ and $\angle COD$ are complementary, by the definition of complementary angles, $m\angle BOC + m\angle COD=90^{\circ}$.

Step2: Use angle - addition

$\angle BOD=\angle BOC+\angle COD$, so by substitution, $m\angle BOD = 90^{\circ}$.

Step3: Recall linear - pair theorem

$\angle AOB$ and $\angle BOD$ form a linear pair. By the linear - pair theorem, $\angle AOB$ and $\angle BOD$ are supplementary, so $m\angle AOB + m\angle BOD=180^{\circ}$.

Step4: Substitute and solve

Substitute $m\angle BOD = 90^{\circ}$ into $m\angle AOB + m\angle BOD=180^{\circ}$, we get $m\angle AOB+90^{\circ}=180^{\circ}$. Then $m\angle AOB = 90^{\circ}$. Since $m\angle AOB=90^{\circ}$ and $m\angle BOD = 90^{\circ}$, $\angle AOB\cong\angle BOD$ (angles with equal measures are congruent).

Answer:

The proof is completed as shown in the steps above to prove $\angle AOB\cong\angle BOD$.