In the given figure, DC = AE = BE, AB // DC. Show that AADE = ACED
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DC = AE = BE
AB II DC
In ▲ AED and ▲ CDE,
∠ AEC = ∠ CDE (alternate interior angles as AB II DC)
AE = CD (given)
ED = DE (common)
Therefore, by SAS congruency
▲ AED and ▲ CDE are congruent ▲s .
pls refer the attachment below for figure :
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