The given figure represents a parallelogram ABCD.

What is the measure of ∠ADC in the given figure?

It is given that ABCD is a parallelogram.

∴AB||CD [In parallelogram, opposite sides are parallel]

∴∠BDC = ∠ABD [Pair of alternate interior angles]

⇒ ∠ABD =65° [∠BDC = 65°]

By applying angle sum property of triangles in ΔOAB, we obtain

∠OAB + ∠AOB + ∠ABO = 180°

⇒ ∠OAB + 95° + 65° = 180°

⇒ ∠OAB = 180° − (95° + 65°) = 180° − 160° 

⇒ ∠OAB = 20°

∴∠BAX = ∠CAX − ∠OAB = 130° − 20° = 110°

∠ADC = ∠BAX [Pair of corresponding angles]

∴∠ADC = 110°