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.
Solution:
We have,
AE = BF = CG = DH = x (say)
BE = CF = DG = AH = y (say)
In ΔAEH and ΔBEF, we have
AE = BF
∠A = ∠B
And AH = BE
So, by SAS congruency criterion, we have
ΔAEH ≃ ΔBFE
⇒ ∠1 = ∠2 and ∠3 = ∠4
But ∠1 + ∠3 = 90° and ∠2 + ∠A = 90°
⇒ ∠1 + ∠3 + ∠2 + ∠A = 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.
