O is a point in the interior of a rhombus ABCD. If OA = OC then prove that DOB is a straight line.

Given: A rhombus ABCD and a point O in it such that OA = OC
To Prove: DOB is a straight line.
Construction: Join OB and OD 
Proof: In ΔAOD and ΔCOD
     AO = CO ...... (given)
     OD = OD ...... (common)
     AD = CD ...... (sides of a rhombus)
... ΔAOD ≅ ΔCOD
.^..∠AOD = ∠COD ...... (i)
Similarly ΔAOB ≅ ΔCOB
... ∠AOB = ∠COB ...... (ii)
Adding eqns (i) and (ii), we get
                             ∠AOD + ∠AOB = ∠COD + ∠COB
But ∠AOD + ∠AOB + ∠COD + ∠COB = 360° ...... (Angles at a point)
  ... ∠AOD + ∠AOB + ∠AOD + ∠AOB = 360°
                    ⇒ 2 (∠AOD + ∠AOB) = 360°
                          ⇒ ∠AOD + ∠AOB = 180° but it is a linear pair 
.^.. OD and OB are in a line
Hence DOB is a straight line.