Question

in the given figure AB=AC, D is a point on AC and E is a point on AB such that AD=ED=EC=BC.prove that angle A: angle B=1:3

Answer 1

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Answer 2

$\underline { \pink{Given :}}$

In ∆ABC AB = AC , D is a point on AC and E is a point on AB such that

AD = ED = EC = BC

$\underline { \pink{To \:Prove :}}$

$\angle A : \angle B = 1 : 3$

$\underline { \pink{Proof :}}$

$i) In \:\triangle AED , \\\: AD = ED \implies \angle {EAD} = \angle {AED}\: ---(1)$

$\blue {( Angles \: opposite \:to \: equal \:sides}\\blue { are \: equal ) }$

$ii) In \:\triangle BEC , \\\: EC = BC \implies \angle {CBE} = \angle {BEC} \: --(2)$

$\angle { BEA} = 180 \degree \: [ Straight \:angle ]$

$iii) In \:\triangle DCE , \\\: DE = EC \implies \angle {DCE} = \angle {CED} \: --(3)$

$\implies \angle { BEC} + \angle { CED} + \angle { AED} = 180\degree$

$\implies \angle { EAD} + \angle { CBE} + \angle { CED} = 180\degree$

$\implies \angle A + \angle B +\angle { CED} = 180\degree$

$\implies \angle A + \angle B + ( 180\degree - 2\angle {EDC} )= 180\degree$

$\blue { Since, (\angle {CED} + 2\angle {EDC} = 180\degree)}$

$\implies \angle A + \angle B - 2 \angle {EDC} = 0$

$\implies \angle A + \angle B - 2\times 2\angle A = 0$

$\implies \angle A + \angle B - 4\angle A = 0$

$\implies \angle B - 3\angle A = 0$

$\implies \angle B = 3\angle A$

$\implies \angle A : \angle B = 1 : 3$

$Hence\: proved$

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