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Upon the sides AB, AC of a AABC are described equilateral triangles ABD aretheir vertices D and E outside the ∆ABC. Prove that BE =DC |
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Answer» We have ABD and ACE are same equilateral triangle , So AB = BD = DA = AC = CE = EAAnd from given information we get our figure , So,AB = ACThenTriangle ABC is a isosceles triangle , So from base angle theorem We get∠ABC =∠ACBAnd∠ABD =∠ ACE = 60° ( As these angles from equilateral triangles ABD and ACE )And ∠CBD =∠ABC +∠ABD ---------- ( 1 )And ∠BCE =∠ACB +∠ACESo,∠BCE =∠ABC +∠ABD ----------- ( 2 ) ( As we know∠ABC =∠ACB And∠ABD =∠ ACE )From equation 1 and 2 , we get∠CBD =∠ BCE ---------- ( 3 )Now in∆CBD and∆BCEBC=BC( Common side )∠CBD =∠ BCE ( From equation 3 )And BD=CE( Given )Hence∆CBD≅∆BCE ( By SAS rule )So,BE=DC ( By CPCT ) ( Hence proved ) |
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