Question

IN THE FIGURE <ABC= 30° ,< EDF= (40-X)° , <ADE= (13X+20)° SHOW THAT BC IS PARALLEL TO DE

ANY MODERATORS PLS ANSWER I WILL MARK AS BRAINLIEST

Answer 1

given that:-
angle ABC=30°
angle EDF=40-x
angle ADE=13x+20

EDF+ADE=180°

40-X+13X+20=180°
60+12X=180
12X=120
X=10°
so angle EDF=30°

now angle EDF=angleABC=30°

both of these are alternate angles
and alternate angles are formed only with parallel lines when transversal cut it .

here || lines are BC and DE
transversal is AF