what is the answer of this question
Answers
Answer:
Step-by-step explanation:
Given : Two Δles ABC and DEF
∠B = ∠E , ∠C = ∠F and BC = EF
RTP : ΔABC ≅ ΔDEF
Case 1 : When AB = DE
AB = DE and ∠B = ∠E
BC = EF
∴ ΔABC ≅ ΔDEF ( SAS CRITERION )
Case 2 : When AB < ED
Construction : Take a point P on ED such that PE = AB . Join FP
In Δ ABC and Δ PEF
AB = PE ( By Supposition)
∠B = ∠E ( Given)
BC = EF ( Given)
∴ ΔABC ≅ ΔPEF ( SAS CRITERION )
⇒ ∠ ACB = ∠ PFE ( CPCT )
∠ ACB = ∠ DFE ( Given)
∴ ∠ PFE = ∠ DFE
This is possible only when P and D Coincide
∴ AB = DE
Thus in Δ ABC and Δ DEF,
AB = DE ( Proved above )
∠ B = ∠E ( Given )
BC = EF (Given )
∴ Δ ABC = Δ DEF ( SAS CRITERION )
Case 3 : When AB > ED
Construction : Take a point M on ED produced such that ME = AB . Join FM
∴ AB = DE
So , ΔABC ≅ ΔDEF
HENCE IN ALL THE THREE CASES ΔABC ≅ ΔDEF
please refer to the below diagram
case 1 , case 2 , case 3 are given respectively
