Question

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 equal to angle ABE and angle EPA equal to angle DPB show that triangle DAP is congruent to triangle EBP and 2 AD equal to BE

Answer 1

Step-by-step explanation:

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