Answer :

AC = CE = EF = AF = 4 cm

∵ B is midpoint of AC

AB = BC = 1/2 AC

⇒ AB = BC = 1/2 × 4 cm

⇒ AB = BC = 2 cm

Similarly,

∵ D is midpoint of AC

CD = DE = 1/2 CE

⇒ CD = DE = 1/2 × 4 cm

⇒ CD = DE = 2 cm

AE is diagonal of ACEF,

∴ By Pythagoras theorem,

AF^{2} + EF^{2} = AE^{2}

⇒ 4^{2} + 4^{2} = AE^{2}

⇒ AE^{2} = 16 + 16

⇒ AE^{2} = 32

⇒ AE= √32

⇒ AE= 4√2 cm

∵ J is midpoint of AE,

∴ AJ = JE = 1/2 AE

⇒ AJ = JE = 1/2 × 4√2 cm

⇒ AJ = JE = 2√2 cm

∵ G is midpoint of AJ,

∴ AG = JG = 1/2 AJ

⇒ AG = JG = 1/2 × 2√2 cm

⇒ AG = JG = √2 cm

∵ BGJH is a square

BG = JH = HB = JG = √2 cm

Also,

HD = HB = √2 cm

And,

IE = IJ = GJ =√2 cm

∵ HD and IE are equal and parallel

∴ HDEI is a parallelogram

⇒ HI = DE = 2 cm

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