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11/07/2021 11:55 am
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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.). Show that
(i) ΔDAP ΔEBP
(ii) AD = BE
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11/07/2021 11:58 am
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.