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In the given figure, ABCD is a parallelogram. AEand BE are angle bisectors of ZA and ZBrespectively. Show that ZAEB =90%.DAB В |
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Answer» ∵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
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.
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