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$AB\perp BC$ & $BD \perp AC$

∴$\measuredangle ABC = 90^{\circ}  $ & $\measuredangle BDA = \measuredangle BDC = 90^{\circ}$

Now, $\angle BAC = 30^{\circ}$

∴$\measuredangle ACB = 180^{\circ} - (90^{\circ}+30^{\circ})$

$\measuredangle ACB = 180^{\circ} - 120^{\circ}$

$\measuredangle ACB = 60^{\circ} $

Given that $CE$ bisects $\angle C$ 

$∴ \measuredangle ECD =  \dfrac{60^{\circ}}{2}= 30^{\circ}  $

Then, $\measuredangle CED = 180^{\circ} - (90^{\circ}+30^{\circ}) = 60^{\circ}$

Answer:

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