Question

In the figure give below DE ∥ BC and AB ∥ MF. Find (i) ∠ADE + ∠MEN (Ii)∠BDE (iii)∠BLE Answer with steps will be marked brainliest

Answer 1

Answer : angle ADE + angle angle MEN = 80. Angle BDE = 140° , angle BLE = 140°

Step by step explanation :

Given ,

DE//BC and AB//MF

When AB//MF DN is a transversal , Then angle ADE and angle MEN are equal .

angle ADE= angle MEN

Now we know that DE//BC then DE//BL similarly DB // ML . DELB is a parallelogram

From the figure , angle DBL= 40°

Then angle DEL = 40°

angle DEL + angle BDE= 180°

40° + BDE. = 180°

angle BDE= 140°

angle BLE = 140°

angle ADE+ angle BDE. = 180°

angle ADE =40°

angle MEN. = 40°

angle ADE+ angle MEN. = 80°

Therefore angle ADE+ angle MEN= 80°

angle BDE = 140°

angle BLE. = 140°

Hope it's correct and helps u ....!!!