--> 0 < [ ∠A, ∠D ] < 90° <--- Acute angles --> Hence, we find a possibility to draw a right-angled Δ
--> Let Figure given define two such Δs ... --> ∠B = 90° || ∠E = 90°
--> However, since a Δ has sum of angles as 180°, we have, 0 < [ ∠A, ∠D ] < 90° { the above condition is met }
--> cos A = [ AB / AC ] = [ DE / DF ] = cos D --> ( i )
However, this also means that :-> --> [ cos² A ] = [ cos² D ] --> [ 1 - sin² A ] = [ 1 - sin² D ] --> sin² A = sin² D
Note :-> 0 < sin A < 1 --> if 0° < A < 90° => sin A = sin D => [ BC / AC ] = [ EF / DF ] --> ( ii )
However, [ sin A ] / [ cos A ] = [ sin D ] / [ cos D ] <-- By ( i ) and ( ii ) => AB / BC = DE / EF => ΔABC ~ ΔDEF <---- [ SSS similarity ] => ∠A = ∠D ^_^
Note --> I didn't use ∠B as stated in the Qn. Rather I replaced ∠B by ∠D
Answers & Comments
--> 0 < [ ∠A, ∠D ] < 90° <--- Acute angles
--> Hence, we find a possibility to draw a right-angled Δ
--> Let Figure given define two such Δs ...
--> ∠B = 90° || ∠E = 90°
--> However, since a Δ has sum of angles as 180°,
we have, 0 < [ ∠A, ∠D ] < 90° { the above condition is met }
--> cos A = [ AB / AC ] = [ DE / DF ] = cos D --> ( i )
However, this also means that :->
--> [ cos² A ] = [ cos² D ]
--> [ 1 - sin² A ] = [ 1 - sin² D ]
--> sin² A = sin² D
Note :-> 0 < sin A < 1 --> if 0° < A < 90°
=> sin A = sin D
=> [ BC / AC ] = [ EF / DF ] --> ( ii )
However, [ sin A ] / [ cos A ] = [ sin D ] / [ cos D ] <-- By ( i ) and ( ii )
=> AB / BC = DE / EF
=> ΔABC ~ ΔDEF <---- [ SSS similarity ]
=> ∠A = ∠D ^_^
Note --> I didn't use ∠B as stated in the Qn. Rather I replaced ∠B by ∠D