Prove this This theorem .........
Answers
Given: Two triangles ABC and DEF such that B = E, C = F
and BC = EF. To prove: ABC DEF
Proof: Case I: If AB = DE then in
ABC and DEF, AB = DE [by supposition] BC = EF [given] and B = E [given]
Thus, ABC DEF [SAS criterion]
Case II: If AB < DE Take a point G on ED such that EG = AB.
Join GF. In ABC and GEF, we have AB = GE [by supposition] BC = EF [given] B = E [given] Thus, ABC GEF [SAS criterion]
ACB = GFE [corresponding parts of congruent triangles are equal]
But ACB = DFE [given] GFE =DFE,
This is only possible when FG coincides with FD or G coincides with D.
AB must be equal to DE and hence, ABC DEF (by SAS) Case III: If AB > ED
With a similar argument (as in case II), we may conclude that ABC DEF (by SAS)
Thus, ABC DEF.
It is all.......
Answer:
Given: Two triangles ABC and DEF such that B = E, C = F
and BC = EF. To prove: ABC DEF
Proof :-
Case I: If AB = DE then in
ABC and DEF, AB = DE [by supposition] BC = EF [given] and B = E [given]
Thus, ABC DEF [SAS criterion]
Case II: If AB < DE Take a point G on ED such that EG = AB.
Join GF. In ABC and GEF, we have AB = GE [by supposition] BC = EF [given] B = E [given] Thus, ABC GEF [SAS criterion]
ACB = GFE [corresponding parts of congruent triangles are equal]
But ACB = DFE [given] GFE =DFE,
This is only possible when FG coincides with FD or G coincides with D.
AB must be equal to DE and hence, ABC DEF (by SAS) Case III: If AB > ED
With a similar argument (as in case II), we may conclude that ABC DEF (by SAS)
Thus, ABC DEF.
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