Answer depends on where we consider the point E to be. If the point lies on line BD, then the above answer by Sukanya is correct. If the point E lies on the line AB, then the answer would be different.

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

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I'm telling that the answer COULD NOT be 65° (or 60°), if the point E lies on the line AB. That is because angle BDE + angle CED = 60°. Proof : consider the point of intersection of lines CE and BD as O. using basic "sum of interior angles of a triangle's law we can find that angle DBA = angle CEB = 60°. Therefore angle EOB = 60. So angle. Hence angle EOD = 180-60= 120°. Now consider the triangle EOD. you will find that angle E + angle D = 180-angle O = 180-120 = 60°. That's how I got angle BDE + angle CED = 60°

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