ABCD is a square. E, F, G and H are points on the sides AB, BC, CD and DA respectively such that AE = BF = CG = DH. Prove that EFGH is a square.

Given, AE = BF = CG = DH

Since, ABCD is a square and AB = BC = CD = AD.

So,

⇒ AB - AE = BC - BF = CD - CG = AD - DH

⇒ EB = FC = GD = AH

Now, in ∆AEH and ∆BFE

⇒ AE = BF [Given]

⇒ AH = EB [Proved]

⇒ ∠A = ∠B [Each 90°]

So, ∆AEH ≅ ∆BFE by S.A.S axiom of congruency

Then, by C.P.C.T we have

⇒ EH = EF and ∠4 = ∠2

In ∆AEH,

⇒ ∠1 + ∠4 + ∠HAE = 180°

⇒ ∠1 + ∠4 + 90° = 180°

⇒ ∠1 + ∠4 = 90°

⇒ ∠1 + ∠2 = 90° [Since, ∠4 = ∠2]

From figure,

⇒ ∠1 + ∠HEF + ∠2 = 180°

⇒ ∠HEF + 90° = 180°

⇒ ∠HEF = ∠E = 90°.

In ∆DGH and ∆CGF

⇒ DH = GC [Given]

⇒ GD = FC [Proved]

⇒ ∠D = ∠C [Each 90°]

So, ∆DGH ≅ ∆CGF by S.A.S axiom of congruency

Then, by C.P.C.T we have

GH = FG

In ∆DGH and ∆AEH

⇒ DH = AE [Given]

⇒ GD = HA [Proved]

⇒ ∠D = ∠A [Each 90°]

So, ∆DGH ≅ ∆AEH by S.A.S axiom of congruency

Then, by C.P.C.T we have

GH = HE

Thus, EF = FG = GH = HE, therefore EFGH is a Rhombus.

∵ One angle of rhombus EFGH is 90° (∠HEF = 90°),

∴ EFGH is a square.

Hence, proved that EFGH is a square.