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Answer :
We have,
AP = BP ...(i)(Since, P is the mid-point of AB)
\(\angle{EPA}\) = \(\angle{DPB}\) ...(ii)(Given)
\(\angle{BAD}\) = \(\angle{ABE}\) ...(iii) (Given)
On adding, \(\angle{EPD}\) on both sides in Equation (ii),
we have,
\(\Rightarrow \) \(\angle{EPA}\) + \(\angle{EPD}\) = \(\angle{DPB}\) + \(\angle{EPD}\)
\(\Rightarrow \) \(\angle{DPA}\) = \(\angle{EPB}\) ...(iv)
Now, In \(\triangle{DAP}\) and \(\triangle{EBP}\) We have,
\(\Rightarrow \) \(\angle{DPA}\) = \(\angle{EPB}\) ...(from (iv)),
\(\Rightarrow \) \(\angle{DAP}\) = \(\angle{EBP}\) ...(Given)
and AP = BP ...(From Eq. (i))
\(\therefore \) \(\triangle{DAP}\) \(\displaystyle \cong \) \(\triangle{EBP}\) ...(By ASA congruency test)
Thus, AD = BE ...(By CPCT)
Hence, proved.