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- In {tex}\triangle{/tex}ABC and {tex}\triangle{/tex}BAC
AD = BC [Given]
{tex} \angle {/tex}DAB = {tex} \angle {/tex}CBA [Given]
AB = AB [Common]
{tex}\therefore \triangle ABD \cong \triangle BAC{/tex} [By SAS congruency] - Since
ΔABD≅ΔBAC
{tex}\therefore {/tex} BD=AC[By C.P.C.T.] - Since
ΔABD≅ΔBAC{tex}\triangle \mathrm{ABD} \cong \triangle \mathrm{BAC}{/tex}
{tex}\therefore{/tex}{tex} \angle {/tex}ABD = {tex} \angle {/tex}BAC [By C.P.C.T.]
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Yogita Ingle 5 years, 1 month ago
(a+b+c)3 - a3-b3-c3 = [(a+b)+c]3 – a3-b3-c3
= (a+b)3+c3+3c(a+b)(a+b+c)-a3-b3-c3
= a3+b3+3ab(a+b)+c3+3c(a+b)(a+b+c)-a3-b3-c3
= a3+b3+c3+3ab(a+b)+3c(a+b)(a+b+c)-a3-b3-c3
= 3ab(a+b)+3c(a+b)(a+b+c) 3(a+b)[ab+c(a+b+c)]
= 3(a+b)[ab+ac+bc+c2] 3(a+b)[a(b+c)+c(b+c)]
= 3(a+b)(b+c)(c+a)
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