Q. 163.5( 8 Votes )

In the given figure, BA AC and DE EF such that BA=DE and BF=DC. Prove that AC=EF.


Answer :

Given : BA AC and DE EF such that BA=DE and BF=DC


To prove: AC = EF


Proof:


In ∆ABC, we have,


BC = BF + FC


And , in ∆DEF,


FD = FC + CD


But, BF = CD


So, BC = BF + FC


And, FD = FC +BF


BC = FD


So, in ∆ABC and ∆DEF, we have,


BAC = DEF … given


BC = FD


AB = DE …given


Thus by Right angle - Hypotenuse- Side property of congruence, we have,


∆ABC DEF


Hence, we know that, corresponding parts of the congruent triangles are equal


AC = EF


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