Show that the bisectors of angle of a parallelogram form a rectangle.
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Qis use as angle ABCD is a llgm as is a Q bisecter of Q A bp is a Q bisecter of Q B therefore Q abs + Qbas =1/2QB+1/2QA
=1/2(QB+QA) =1/2 *180(Q A and QD are interior Q on the same side of the transversal)
=90 =Qabs +Qbas+Qasb =180 (angle sum property)
90+Qasb = 180
Qasb=180 - 90 = 90 qpsr = 90 (vertically opposite Q)
one of Q is 90 that is a rectangle (property of a rectangle
=1/2(QB+QA) =1/2 *180(Q A and QD are interior Q on the same side of the transversal)
=90 =Qabs +Qbas+Qasb =180 (angle sum property)
90+Qasb = 180
Qasb=180 - 90 = 90 qpsr = 90 (vertically opposite Q)
one of Q is 90 that is a rectangle (property of a rectangle
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