Prove that in an isosceles trapezoid, adjacent angles at the parallel sides are equal?

1 Answer
Oct 18, 2017

Please see below.

Explanation:

Let us consider the isosceles trapezoid #ABCD#, where #BC#||#AD# and #AB=CD# as shown below, in which we have drawn #CE#||#AB#.

As #CE#||#AB#, we have #m/_BAE=m/_CED#
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As #AB#||#CE# and #AE#||#BC#, #ABCE# is a parallelogram and hence #AB=CE# and , but as #AB=CD#, we have #CD=CE#.

Therefore #DeltaCDE# is an isosceles triangle and hence #m/_CED=m/_CDE#.

But as #m/_BAE=m/_CED#, we have #m/_BAE=m/_CDE# or #m/_BAD=m/_ADC#

Further as #AD#||#BC#, #/_ABC# is supplementary to #/_BAD# and similarly #/_ADC# and #/_BCD# are supplementary.

As #/_ABC# and #/_BCD# are supplementary of equal angles, they too are equal.

#Q.E.D.#

Note #-# I have not used #E# as a midpoint of #AD#.