In a triangle ABC, the side BC is extended up to D such that CD = AC. If ∠BAD = 109° and ∠ACB = 72° then the value of ∠ABC is

According to question,
Given :

   ∠BAD = 109°
   ∠ACB = 72°
∴ ∠ACD = 180° - 72°
   ∠ACD = 108°
∴      AC = CD
   ∠CAD = ∠CDA
In ΔCDA
∠CAD + ∠CDA + ∠DCA = 180°
2∠CAD + 108° = 180°
2∠CAD = 180° -108°
2∠CAD = 72°
∠CAD = $$\frac{{{{72}^ \circ }}}{2}$$
∠CAD = 36°
∴ ∠CAB = 109° - 36°
∠CAB = 73°
In ΔABC
∠ABC + ∠ACB + ∠CAB = 180°
∠ABC + 72° + 73° = 180°
∠ABC + 145° = 180°
∠ABC = 180° - 145°
∠ABC = 35°