In the adjoining figure, AB || DC. CE and DE bisects ∠BCD and ∠ADC respectively. Prove that AB = AD + BC.

Given,

DE bisects ∠D.

∴ ∠EDA = ∠EDC = $\dfrac{∠D}{2}$.

AB || CD and DE cuts these parallel lines.

∠DEA = ∠EDC = $\dfrac{∠D}{2}$.

In △ADE,

∠ADE = ∠DEA = $\dfrac{∠D}{2}$.

Hence, AD = AE.

Given,

CE bisects ∠C.

∴ ∠ECB = ∠ECD = $\dfrac{∠C}{2}$.

AB || CD and CE cuts these parallel lines.

∠CEB = ∠ECD = $\dfrac{∠C}{2}$.

In △BCE,

∠BEC = ∠BCE = $\dfrac{∠C}{2}$

Hence, BE = BC.

⇒ AB = AE + BE = AD + BC

⇒ AB = AD + BC.

Hence, proved that AB = AD + BC.