question_answer

   Explain, why\[\Delta ABC\cong \Delta FED\].  

Answer:

Since, (i)            \[\angle A=\angle F\]                   (given)                                                                                       (ii)                     BC = DE           (given) \[\angle B=\angle E=90{}^\circ \]           (given) Hence, \[\Delta ABC\cong \Delta FED\] by ASA criterion of congruency \[\therefore \]                  \[\angle C=\angle D\] (As Z\[\angle \]A=\[\angle \]F, \[\angle \]B=\[\angle \]E \[\therefore \]3rd \[\angle \]C = 3rd \[\angle \]D) So, by ASA criterion of congruence the result holds.