Show that parallelograms which are on the same base and in the same parallels equal one another? This is called Euclid Proposition I.35...

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Answer 1 · 2018-01-28T12:45:32

As proved below.

To prove Area of parallelogram ABCD = Parallelogram BEFC.

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Refer diagram

#AD = BC, EF = BC#

#:. AD = EF#

Consider triangles ABE and DCF

#AB = DC, BE = CF#

#AE = AD + DE# & #DF = DE + EF#

We have already proved, AD = EF and DE is common.

Hence #Delta ABE & DCF# congruent.

Area #ABCF = cancel(ABE) + EBCF = ABCD + cancel(DCF)#

Since ABE = DCF#

Therefor Parallelogram ABCD = Parallelogram EBCF#