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Step-by-step explanation:
i)
It is given that ∠EPA=∠DPB
Now,
∠EPA+∠DPE=∠DPB+∠DPE
Therefore,
∠DPA=∠EPB
In ΔEBP and ΔDAP,
∠EBP=∠DAP (given)
BP=AP (P is midpoint of AB)
∠EPB=∠DPA [proved above]
By ASA criterion of congruence,
ΔEBP≅ΔDAP
ii)
Since ΔEBP≅ΔDAP
AD=BE (using CPCT
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