In the given figure, in ΔABC it is given that ∠B = 40°and ∠C = 50°, DE || BC, and EF || AB Find: (i) ∠ADE + ∠MEN (ii) ∠BDE and (iii) ∠BFE

Since DE || BC and AB acts as transversal.

So, ∠ADE = ∠ABC {corresponding angles}

since ∠ABC = 40°

So, ∠ADE = 40°

Since EF || AB and DN acts as transversal.

So, ∠ADE = ∠MEN {corresponding angles}

∠MEN = 40°

Hence, ∠ADE + ∠MEN = 80°

(ii) 140°

Since AB represents a straight line. Sum of angles in line AB = 180°

So, ∠BDE + ∠ADE = 180°

since, ∠ADE = 40°

So, ∠BDE = 140°

(iii) 140°

Since DE || BC and FM acts as transversal.

So, ∠EFC = ∠ MEN = 40°

And BC represents a straight line. Sum of angles in line BC = 180°

= ∠EFC + ∠BFE = 180°

= ∠BFE = 140°