In the given figure, AB = AC, D is a point on AC and E an AB such that AD = ED = EC = BC, The ratio $\angle{BAC} : \angle{ABC} =$ ___

1 : 2

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1 : 3

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2 : 1

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3 : 1

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Solution

The correct option is B
1 : 3

AD = DE = EC = BC (given)

$∠ D A E = ∠ A E D = x (s a y)$

$∴ ∠ E D C = 2 x . . .$ (exteriar angle theorem)

$∴ ∠ E C D = 2 x . . .$ (ED = EC, base angles of an isosceles $Δ$ )

$∴ ∠ C E B = x + 2 x$ (exteriar angle theorem)

= 3x

$∴ ∠ A B C = ∠ C E B = 3 x$ (base angles of an isosceles $Δ$ )

$∠ B A C : ∠ A B C = x : 3 x$

= 1 : 3

Hence(B)

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In the given figure, AB = AC, D is a point on AC and E an AB such that AD = ED = EC = BC, The ratio \(\angle{BAC} : \angle{ABC} =\) ___