9. ABCD is a parallelogram. AE and AF are bisectors of
interior and exterior angles respectively at A and BE
and BF are bisectors of interior and exterior angles
respectively at B. Show that AFBE is a parallelo-
gram.
Answers
Step-by-step explanation:
∵AEisbisectorofanangle∠AB
∴∠DAE=∠EAB..............(i)
∴BEisbisectofangle∠CBA
∴∠CBE=∠EBA...............(ii)
DA∥BC
∠DAB+∠CBA=180
∠DAE+∠EAB+∠CBE+∠EBA=180
fromeq(i)and(ii)
∠EAB+∠EAB+∠EAB+∠EAB=180
0
2(∠EAB+∠EBA)=180
0
∠EAB+∠EAB=90
0
..............(iii)
InΔAEB
∠AEB+∠EAB+∠EBA=180
0
(propertyofΔ)
fromeq
n
...(iii)
∠AEB+90=180
0
∠AEB=90
0
∴∠E=90
0
........(iv)
Similarly,
∠F=90
0
.........(v)
∵AFisbisectof∠BAM
∴∠BAF=∠MAF......(vi)
∠DAB+∠BAM=180
o
(linearpair)
∠DAB+EAB+∠BAF+∠MAF=180
0
Fromeq
n
(i)and(vi)
∠EAB+∠EAB+∠BAF+∠BAF=180
0
⇒∠EAB+∠BAF=90
0
⇒∠A=90
0
..........(vii)
SimilarlyBFbisect∠ABNandwewillobtain
∠B=90
0
.........(vii)
Since∠A=∠B=90
0
and∠E=∠F=90
0
oppositeanglesareequalhenceAFBEisaparallelogram.
AEisbisectorofanangle∠AB
∴∠DAE=∠EAB..............(i)
∴BEisbisectofangle∠CBA
∴∠CBE=∠EBA...............(ii)
DA∥BC
∠DAB+∠CBA=180
∠DAE+∠EAB+∠CBE+∠EBA=180
fromeq(i)and(ii)
∠EAB+∠EAB+∠EAB+∠EAB=180
0
2(∠EAB+∠EBA)=180
0
∠EAB+∠EAB=90
0
..............(iii)
InΔAEB
∠AEB+∠EAB+∠EBA=180
0
(propertyofΔ)
fromeq
n
...(iii)
∠AEB+90=180
0
∠AEB=90
0
∴∠E=90
0
........(iv)
Similarly,
∠F=90
0
.........(v)
∵AFisbisectof∠BAM
∴∠BAF=∠MAF......(vi)
∠DAB+∠BAM=180
o
(linearpair)
∠DAB+EAB+∠BAF+∠MAF=180
0
Fromeq
n
(i)and(vi)
∠EAB+∠EAB+∠BAF+∠BAF=180
0
⇒∠EAB+∠BAF=90
0
⇒∠A=90
0
..........(vii)
SimilarlyBFbisect∠ABNandwewillobtain
∠B=90
0
.........(vii)
Since∠A=∠B=90
0
and∠E=∠F=90
0
oppositeanglesareequalhenceAFBEisaparallelogram.