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In △ABC, AB=AC
Thus, ∠BAC=∠BCA ....(Isosceles triangle property).
Also, ∠BDC=∠BAC=25
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....(Angles in same segment).
In △BDC,
by angle sum property, the sum of angles =180
o
⟹ ∠CBD+∠BCA+∠ACD+∠BDC=180
⟹ ∠ACD=180−75=105
∘
.
Now, in cyclic quadrilateral ACDE,
∠ACD+∠AED=180 ....(Opposite angles of cyclic quadrilateral)
⟹ ∠AED=180−105
⟹ ∠AED=75
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