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In the given figure, $AB = BC = CD$ . If $∠ BAC = 25^{\circ}$ , then which of the following is the value of $∠ AED$ ?

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50^{\circ}

60^{\circ}

65^{\circ}

75^{\circ}

answer is D.

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Detailed Solution

Given,

AB = BC = CD

and

∠ BAC = 25^{\circ}

Here,
In

△ ABC

AB = AC

Thus,

∠ BAC = ∠ BCA

....(Isosceles triangle property).
Also,

∠ BDC = ∠ BAC = 25^{\circ}

....(Angles in the same segment).
In

△ BDC

,
By angle sum property, the sum of all interior angles in a triangle is

180^{o}

⟹

∠ CBD + ∠ BCA + ∠ ACD + ∠ BDC = 180 °

⟹

∠ ACD = 180 ° - (∠ CBD + ∠ BCA + ∠ BDC)

⟹

∠ ACD = 180 ° - 75 °

⟹

∠ ACD = 105^{\circ}

Now, in cyclic quadrilateral

ACDE

∠ ACD + ∠ AED = 180

....(Opposite angles of cyclic quadrilateral)

⟹

∠ AED = 180 ° - 105 °

⟹

∠ AED = 75^{\circ}

Hence, the measure of

∠ AED

75^{\circ}

.
Therefore, option 4 is correct.

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