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In the figure, ABCD is a square and CDE is an equilateral triangle. Prove that :



(i) AE = BE                     (ii) ∠ E B C = 15 °

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 https://www.meritnation.com/ask-answer/question/abcd-is-a-square-and-on-the-side-dc-an-equilateral-triangle/quadrilaterals/6533714
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(i) AD=BC (all side of a square are equal)1
    DE=CE (equilateral triangle)2
   adding 1 and 2
   i.e.
    AD+DE=BC+CE(equals when added to equals the wholes are equal)
therefore, AE=BE
 
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Hello Aashish,
(i) Take triangles ADE and BCE
AD = BC (sides of square)
DE = CE (sides of equilateral)
<ADE = 90+60 = 150
<BCE = 90 + 60 = 150
So <ADE = <BCE
SAS confirms that both triangles are congruent
Hence AE = BE [proved]

(ii) BCE is an isosceles triangle as sides BC = CE
​Hence angles <EBC + <BEC + <BCE = 180
<BCE  = 150
Being isosceles, 2 <EBC = 180-150 = 30
SO <EBC = 15o
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