In figure, bisectors of ∠ B and ∠ D of a quadrilateral ABCD meet CD and AB produced at P and Q respectively. Prove that ∠ P +∠ Q=(∠ ABC + ∠ ADC).
          

Given: ABCD is a quadrilateral. The bisectors of ∠ B and ∠ D meet CD and AB produced at P and Q.
To Prove: ∠ P + ∠ Q = (∠ ABC + ∠ ADC).
Proof:
In Δ PBC,
        ∠ P = 180° - (∠ C + ∠ PBC) [ Angle sum property of a triangle]
But ∠ PBC = (Given)
     ∴ ∠ P = 180° - (∠ C+) …………………(i)
|||y, in Δ ADQ,
∠ Q = 180° - (∠ A + )…………………..(ii)
Adding (i) and (ii),
∠ P + ∠ Q = 180° - (∠ C + ) + 180° - (∠ A + )
Adding and subtracting ( + )
∠ P + ∠ Q = 360° - (∠ A + ∠ B + ∠ C + ∠ D) ++
=360 - 360° + +
= +
= +
= (∠ ABC + ∠ ADC)
∴ ∠ P + ∠ Q = (∠ ABC + ∠ ADC)