In figure, AP and DP are the bisectors of two adjacent angles A and D of a quadrilateral ABCD. Prove that 2∠APD = ∠B + ∠C.
      

Given: ABCD is a quadrilateral. AP and DP are the bisectors of ∠A and ∠D intersecting at P.
To prove: 2∠APD = ∠B + ∠C
Proof:
From quadrilateral ABCD, ∠A + ∠B + ∠C + ∠D = 360° [Sum of the four angles of a quadrilateral is 360° ]
       ഊ + ∠D = 360° - (∠ B + ∠ C) ……………………………..(1)
From Δ APD, ∠ APD + ∠ PAD + ∠ ADP = 180° [Sum of the three angles of a triangle is 180° ]
∠APD ++ = 180° [Since + ∠PAD = , ∠ADP = ]
∴ ∠APD = 180° --
            = 180° -(∠A + ∠D)
            =
\\ 2∠APD = 360° -[360° - (∠B + ∠C)]   [from (1)]
             = 360° -360° + ∠B + ∠C
             = ∠B + ∠C.
∴ 2∠APD = ∠B + ∠C.