Answer :

D is the centre of both the circles.

DG is a perpendicular drawn on EF(HI).

We have to prove that, EH = IF.

In ΔDHG and ΔDIG we have,

DH = DI [radius of the same circle]

DG is the common side.

DG is perpendicular on HI.

∴ ∠DHG = ∠DIG = 90°

∴ ∆DHG≅∆DIG [SAS congruency]

∴ GH = GI …………… (1)

In ΔDEG and ΔDFG we have,

DE = DF [radius of the same circle]

DG is the common side.

DG is perpendicular on EF.

∴ ∠DEG = ∠DFG = 90°

∴ ∆DEG≅∆DFG [SAS congruency]

∴ GE = GF …………… (2)

From, (2) – (1) we get,

⇒ GE – GH = GF – GI

⇒ EH = IF

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