In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE.


It is given in the question that:

AC = AE


AB = AD and,


BAD = EAC


To show: BC = DE


Proof:


BAD = EAC (Adding DAC both sides)


BAD + DAC = EAC + DAC


BAC = EAD


In


AC = AE (Given)


BAC = EAD


AB = AD (Given)


Therefore,


By SAS congruence,



BC = DE (By c.p.c.t)


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