AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \(\angle{BAD}\) = \(\angle{ABE}\) and \(\angle{EPA}\) = \(\angle{DPB}\)(see figure). Show that
(i) \(\triangle{DAP}\) \(\displaystyle \cong \) \(\triangle{EBP}\)
(ii) AD = BE

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.