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Sia ? 3 years, 2 months ago
Given BA || ED and BC || EF.
To show {tex}\angle{/tex}ABC = {tex}\angle{/tex}DEF.
Construction Draw a ray EP opposite to ray ED.
Proof In the figure, BA || ED or BA || DP
{tex}\therefore{/tex} {tex}\angle{/tex}ABP = {tex}\angle{/tex}EPC [corresponding angles]
{tex}\Rightarrow{/tex} {tex}\angle{/tex}ABC = {tex}\angle{/tex}EPC ....(i)
Again, BC || EF or PC || EF
{tex}\therefore{/tex} {tex}\angle{/tex}DEF = {tex}\angle{/tex}EPC [corresponding angles] ....(ii)
From eqs.(i) and (ii),
{tex}\angle{/tex}ABC = {tex}\angle{/tex}DEF
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Sia ? 3 years, 2 months ago
(a−2b−3c)2
According to the equation ,
(a+b+c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
So we get,
Using the formula we can it write it as
(a−2b−3c)2 = a2+(−2b)2+(−3c)2+2a(−2b)+(2(−2b)(−3c)+2(−3c)a
On further calculation we get,
(a−2b−3c)2 = a2 + 4b2 + 9c2 − 4ab + 12bc − 6ac
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