∵AEisbisectorofanangle∠AB∴∠DAE=∠EAB..............(i)∴BEisbisectofangle∠CBA∴∠CBE=∠EBA...............(ii)DA∥BC∠DAB+∠CBA=180∠DAE+∠EAB+∠CBE+∠EBA=180fromeq(i)and(ii)∠EAB+∠EAB+∠EAB+∠EAB=18002(∠EAB+∠EBA)=1800∠EAB+∠EAB=900..............(iii)InΔAEB∠AEB+∠EAB+∠EBA=1800(propertyofΔ)fromeqn...(iii)∠AEB+90=1800∠AEB=900∴∠E=900........(iv)Similarly,∠F=900.........(v)∵AFisbisectof∠BAM∴∠BAF=∠MAF......(vi)∠DAB+∠BAM=180o(linearpair)∠DAB+EAB+∠BAF+∠MAF=1800Fromeqn(i)and(vi)∠EAB+∠EAB+∠BAF+∠BAF=1800⇒∠EAB+∠BAF=900⇒∠A=900..........(vii)SimilarlyBFbisect∠ABNandwewillobtain∠B=900.........(vii)Since∠A=∠B=900and∠E=∠F=900oppositeanglesareequalhenceAFBEisaparallelogram.