7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see Fig. 7.22). Show that (i) ΔDAP ≅ ΔEBP (ii) AD = BE
Solutions: In the question, it is given that P is the mid-point of line segment AB. Also, ∠BAD = ∠ABE and ∠EPA = ∠DPB (i) It is given that ∠EPA = ∠DPB Now, add ∠DPE on both sides, ∠EPA +∠DPE = ∠DPB+∠DPE This implies that angles DPA and EPB are equal i.e. ∠DPA = ∠EPB Now, consider the triangles DAP and EBP. ∠DPA = ∠EPB AP = BP (Since P is the mid-point of the line segment AB) ∠BAD = ∠ABE (As given in the question) So, by ASA congruency, ΔDAP ≅ ΔEBP. (ii) By the rule of CPCT, AD = BE.
