Question

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 parallelogram.

Answer 1

$\begin{array}{c}\because AEisbi\sec torofanangle\angle AB\\ \therefore \angle DAE=\angle EAB..............\left(i\right)\\ \therefore BEisbi\sec tofangle\angle CBA\\ \therefore \angle CBE=\angle EBA...............\left(ii\right)\\ DA\parallel BC\\ \angle DAB+\angle CBA=180\\ \angle DAE+\angle EAB+\angle CBE+\angle EBA=180\\ fromeq\left(i\right)and\left(ii\right)\\ \angle EAB+\angle EAB+\angle EAB+\angle EAB={180}^0\\ 2(\angle EAB+\angle EBA)={180}^0\\ \angle EAB+\angle EAB={90}^0..............\left(iii\right)\\ In\Delta AEB\\ \angle AEB+\angle EAB+\angle EBA={180}^0\left(propertyof\Delta \right)\\ frome{q}^n...\left(iii\right)\\ \angle AEB+90={180}^0\\ \angle AEB={90}^0\\ \therefore \angle E={90}^0........\left(iv\right)\\ Similarly,\\ \angle F={90}^0.........\left(v\right)\\ \because AFisbi\sec tof\angle BAM\\ \therefore \angle BAF=\angle MAF......\left(vi\right)\\ \angle DAB+\angle BAM={180}^o\left(linearpair\right)\\ \angle DAB+EAB+\angle BAF+\angle MAF={180}^0\\ Frome{q}^n\left(i\right)and\left(vi\right)\\ \angle EAB+\angle EAB+\angle BAF+\angle BAF={180}^0\\ \Rightarrow \angle EAB+\angle BAF={90}^0\\ \Rightarrow \angle A={90}^0..........\left(vii\right)\\ SimilarlyBFbi\sec t\angle ABNandwewillobtain\\ \angle B={90}^0.........\left(vii\right)\\ Since\angle A=\angle B={90}^0\\ and\angle E=\angle F={90}^0\\ oppositeanglesareequalhenceAFBEisaparalle\log ram.\end{array}$