Question

hello everyone...✌✌
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pls solve the above question...

Answer 1

Answer: In ΔDAP and ΔEBP
∠BAD=∠ABE∠BAD=∠ABE (given)
∠EPA=∠DPB∠EPA=∠DPB (given)
So, ∠EPA+∠EPD=∠DPB+∠EPD∠EPA+∠EPD=∠DPB+∠EPD
Or, ∠DPA=∠EPB∠DPA=∠EPB
AP=PBAP=PB (P is midpoint of AB)
So, by ASA rule ΔDAP≅ΔEBPΔDAP≅ΔEBP
So, 

Answer 2

❤️❤️HEY MATE❤️❤️

⭐️ANSWER⭐️

✔️About the question====)
✴️Given-
DAP =EPP
➡️AD=BE

☑️SOLUTION
we have -
EPA=DPB
=)EPA+DPE=DPE+DPE
DPA=EPB ...1

Now in EPB and DAP we have
EPB = DPA(from ..1)
BP=AP(given)
And , EPB =DAP(given)
so , by ASA criterion of congruence , we have

EPB=DAP
BE=AD(that is AD =BE)

=) BY CPCT

❤shinchan#