116 MATHEMATICS
Case (i) : Let AB = DE (see Fig. 7.12).
Now what do you observe? You may observe that
AB = DE (Assumed)
✄B =✄E (Given)
BC = EF (Given)So, ✂ABC ☎✂DEF (By SAS rule)
Fig. 7.12Case (ii) : Let if possible AB > DE. So, we can take a point P on AB such that
PB = DE. Now consider ✂PBC and ✂DEF (see Fig. 7.13).
Fig. 7.13Observe that in ✂PBC and ✂DEF,
PB = DE (By construction)
✄B =✄E (Given)
BC = EF (Given)So, we can conclude that:
✂PBC ☎✂DEF, by the SAS axiom for congruence.