In Figure, BA AC, DE DF such that BA = DE and BF = EC. Show that ΔABC ΔDEF.


Given: BA AC, DE DF such that BA = DE and BF = EC.

In ΔABC and ΔDEF


BA = DE (given)


BF = EC(given)


A = D (both 90°)


BC = BF + FC


EF = EC + FC = BF + FC ( EC = BF)


EF = BC


Hence, ΔABC ΔDEF (by RHS)


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