Sir please help me to solve the ncert class 9^{th} solution of chapter triangles.How I solve this question of exercise 7.1 question number 7. Find the simplest and easiest solution of this question , also give me the best solution of this question.

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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 BAD = ABE and EPA = DPB (see Fig. 7.22). Show that (i) ΔDAP ΔEBP (ii) AD = BE . Q.7

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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.