Help with Geometry Proofs

I suck at doing proofs. Could you help me.

/_AOB and /_BOC are a linear pair;
m/_AOB=m/_BOC
Prove: /_AOB and /_BOC are right angles

I just don't get it :( (/_=the angle sign)

Start with what you are given:
1. ∠AOB and ∠BOC are a linear pair (Given)
2. m∠AOB = m∠BOC (Given)

Now remember what the definition of a linear pair of angles means. The definition is:
m∠AOB + m∠BOC = 180º

Hence for our step 3 of the proof we are using the fact that ∠AOB and ∠BOC are a linear pair to write:
3. m∠AOB + m∠BOC = 180º (By definition of a linear pair)

We also know from step 2 that m∠AOB = m∠BOC, so plug in m∠BOC for m∠AOB is step 3:
4. m∠BOC + m∠BOC = 180º

Combine like terms to get:
5. 2*m∠BOC = 180º

Divide both sides by the number 2:
6. m∠BOC = 90º

So ∠BOC is a right angle. Now going back to step 3 and plugging in m∠BOC = 90º, we have:
7. m∠AOB + 90º = 180º

Subtract 90º from both sides:
8. m∠AOB = 90º

So ∠AOB is a right angle as well. Thus we have shown both ∠AOB and ∠BOC are right angles.




Or in short:
1. ∠AOB and ∠BOC are a linear pair (Given)
2. m∠AOB = m∠BOC (Given)
3. m∠AOB + m∠BOC = 180º (By definition of a linear pair)
4. m∠BOC + m∠BOC = 180º (By steps 2 and 3)
5. 2*m∠BOC = 180º (Combine like terms)
6. m∠BOC = 90º (Divide through by number 2)
7. m∠AOB + 90º = 180º (By steps 3 and 6)
8. m∠AOB = 90º (Subtract 90º from both sides)
9. ∠AOB and ∠BOC are right angles (By steps 6 and 8)

(1) ∠AOB and ∠BOC are...........Given
......a linear pair

(2) ∠AOB = ∠BOC .....................Given

(3) ∠AOB + ∠BOC = 180°..........Definition of LINEAR PAIRed angles

(4) ∠AOB + ∠AOB = 180°........Substitution from (2)

(5) 2 ∠AOB = 180°......................Angle Addition Property

(6) ∠AOB = 180° / 2....................Division

(7) ∠AOB = 90°...........................Simplify

(8) ∠AOB = ∠BOC .....................Given

(9) ∠BOC = 90°............................Transitiv…‡ Property