In that figure, if `Delta ABE ~= Delta ACD`, show that `Delta ADE ~= Delta ABC`.
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`Delta ABE~= Delta ACD` (give)
`:. AE= AD” “…(i)` [cpct]
and `AB =AC” “…(ii)` [ cpct]
In `Delta ADE and Delta ABC`, we have
`angle DEA= angle BAC` (common)
and `(AD)/(AB)=(AE)/(AC)” “` [ using (i) and (ii)]
`:. Delta ADE ~ Delta ABC`[ by SAS-similarity]