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Sia ? 2 years, 5 months ago
Consider the triangles AEB and CDB
{tex}\angle{/tex}EBA = {tex}\angle{/tex}DBC ... (Common angle) ... (i)
Further, we have:
{tex}\angle{/tex}BEA = 180 - y
{tex}\angle{/tex}BDC = 180 - x
Since x = y, we have:
180 - x = 180 - y
{tex}\angle{/tex}BEA = {tex}\angle{/tex}BDC ... (ii)
AB = CB ... (Given) ... (iii)
From (i), (ii) and (iii), we have:
{tex}\triangle{/tex}BDC{tex}\cong{/tex}{tex}\triangle{/tex}BEA ... (AAS criterion)
{tex}\therefore{/tex} AE = CD (CPCT)
Hence, proved.
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