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# 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.