Question

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

Answer 1

Answer:

In a quadrilateral ABCD, AB is parallel to CD. DE and CE bisects Angle ADC and Angle BCD respectively. Prove that AB is equal to the sum of AD and BC.​

DE Bisect ∠D

=> ∠EDA = ∠EDC = ∠D/2

AB ║ CD  & DE is line cutting these parallel lines

=> ∠DEA  = ∠EDC  = ∠D/2

in Δ ADE  

∠EDA  & ∠DEA = ∠D/2

Hence AD = AE

Similarly

in Δ BCE  

∠ECB  = ∠CEB = ∠C/2

=> BE = BC

AB = AE + BE = AD + BC

=> AB = AD + BC

QED

Proved

Answer 2

Step-by-step explanation:

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