Given two triangles ABC and DEF where AC = DF, AB = DE, angle B is a right angle, angle E is obtuse, and angle F < angle C, is BC greater than, equal to, or less than EF. Support your answer with a proof.
Given: ABC be any triangle right angled at B. DEF be another triangle such that AB = DE, AC = DF and ∠DEF is obtuse.
To Prove: BC > EF or BC = EF or BC < EF whichever is true
Construction: Apply ΔDEF on ΔABC in the following manner: put D on A and DF along AC. As DF = AC (given) hence F falls on C. Join BE.
Proof:
In ΔBAE; AB = AE (since AE = DE and DE = AB).
Hence ∠ABE = ∠AEB
Since ∠ABC < ∠AEC (∠ABC = 90; ∠AEC is obtuse).
Therefore ∠ABC - ∠ABE < ∠AEC - ∠AEB
That is in ΔBCE
∠EBC < ∠CEB.
As side opposite to the bigger angle is bigger, BC > EF. (proved)
Given: ABC be any triangle right angled at B. DEF be another triangle such that AB = DE, AC = DF and ∠DEF is obtuse.
To Prove: BC > EF or BC = EF or BC < EF whichever is true
Construction: Apply ΔDEF on ΔABC in the following manner: put D on A and DF along AC. As DF = AC (given) hence F falls on C. Join BE.
Proof:
In ΔBAE; AB = AE (since AE = DE and DE = AB).
Hence ∠ABE = ∠AEB
Since ∠ABC < ∠AEC (∠ABC = 90; ∠AEC is obtuse).
Therefore ∠ABC - ∠ABE < ∠AEC - ∠AEB
That is in ΔBCE
∠EBC < ∠CEB.
As side opposite to the bigger angle is bigger, BC > EF. (proved)
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