ABCD is a rhombus. AB is produced to F and BA is produced to E such that AB = AE = BF. Then :

ABCD is a rhombus and AB = AE = BF

=> AD = AE and BC = BF

=> $$\angle$$AED = $$\angle$$ADE and $$\angle$$BCF = $$\angle$$BFC

Let $$\angle$$DAB = $$2\theta$$ and $$\angle$$ABC = $$2\alpha$$

Using exterior angle property, we get :

=> $$\angle$$AED = ADE = $$\theta$$

and $$\angle$$BCF = $$\angle$$BFC = $$\alpha$$

Also, in ABCD

=> $$\angle$$DAB + $$\angle$$ABC = 180

=> $$2\theta + 2\alpha = 90$$

=> $$\theta + \alpha = 90$$

Now, in $$\triangle$$GEF

=> $$\angle G$$ + $$\theta + \alpha$$ = 180

=> $$\angle G$$ = 180 - 90 = 90

=> $$ED \perp CF$$