Discussion

Explanation:

Let us draw EP perpendicular to AB and join EA.

Let AP = x

In ∆BEP, ∠EBP = ∠BEP = 45°
∴ BP = EP = 3 + x

In ∆AEP
⇒ AE² = AP² + PE²
⇒ 6² = (x)² + (3 + x)²
⇒ 36 = x² + 9 + x² +  6x
⇒ 2x² + 6x - 27 = 0
⇒ x = $\frac{-6+6\sqrt{7}}{4}$ = $\frac{3\sqrt{7}-3}{2}$

⇒ EB = √2(3 + x) = $\sqrt{2}\left(3+\frac{3\sqrt{7}-3}{2}\right)$ = $\frac{3\sqrt{7}+3}{\sqrt{2}}$

⇒ ED = EB - DB =  $\frac{3\sqrt{7}+3}{\sqrt{2}}$ - 3 = $\frac{3\sqrt{7}-3\sqrt{2}+3}{\sqrt{2}}$

Hence, option (b).

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