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.
You need to login to perform this action.
You will be redirected in
3 sec