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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
3Thank You