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Question
ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.
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Solution 1
We have
AE = BF = CG = DH = x (say)
∴ BE = CF = DG = AH = y (say)
In Δ ' s AEH and BEF , we have
AE = BF
`∠`A = `∠`B
And AH = BE
So, by SAS configuration criterion, we have
ΔAEH ≅ ΔBFE
⇒ `∠`1 = `∠`2 and `∠`3 = `∠`4
But `∠`1+ `∠`3 = 90° and `∠`2 + `∠`4 = 90°
⇒ `∠`1+ `∠`3+ `∠`2 + `∠`4 = 90° + 90°
⇒ `∠`1+ `∠`4 + `∠`1+ `∠`4 = 180°
⇒ 2 (`∠`1+ `∠`4) = 180°
⇒ `∠`1+ `∠`4 = 90°
HEF = 90°
Similarly we have `∠`F = `∠`G = `∠`H = 90°
Hence, EFGH is a square
Solution 2
We have
AE = BF = CG = DH = x (say)
∴ BE = CF = DG = AH = y (say)
In Δ ' s AEH and BEF , we have
AE = BF
`∠`A = `∠`B
And AH = BE
So, by SAS configuration criterion, we have
ΔAEH ≅ ΔBFE
⇒ `∠`1 = `∠`2 and `∠`3 = `∠`4
But `∠`1+ `∠`3 = 90° and `∠`2 + `∠`4 = 90°
⇒ `∠`1+ `∠`3+ `∠`2 + `∠`4 = 90° + 90°
⇒ `∠`1+ `∠`4 + `∠`1+ `∠`4 = 180°
⇒ 2 (`∠`1+ `∠`4) = 180°
⇒ `∠`1+ `∠`4 = 90°
HEF = 90°
Similarly we have `∠`F = `∠`G = `∠`H = 90°
Hence, EFGH is a square
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