question_answer

In the given figure, \[\Delta \]ABC is an isosceles triangle in which AB = AC. If AB and AC are produced to D and E, respectively such that BD = CE. Prove that BE = CD.

Answer:

Since, AB = AC                      [given] BD = CE                       [given] \[\therefore \]      AB + BD = AC + CE [adding the two] AD = AE Now in DADC and DAEB AD = AE AC = AB                      [given] \[\angle A\text{ }=\angle A\]                    [Common] So, by S.A.S. congruency we have \[\Delta ADC\cong \Delta AEB\] by c.p.c.t., CD = BE 3