Question : In $\triangle ABC, DE \parallel BC$ in such a way that $A-D-B$ and $A-E-C$ are equal. If $ \angle A C B=40°$, then $\angle D A E+\angle ADE =$ ___________.
Option 1: 240°
Option 2: 120°
Option 3: 140°
Option 4: 230°
Correct Answer: 140°
Solution :
Given,
A triangle ABC in which DE || BC.
Here, we have a triangle ABC in which DE $\parallel$ BC
⇒ The $\angle$ACB = $\angle$AED = 40° (corresponding angles as DE$\parallel$BC and AC is a transversal line which intersects them)
Now, consider the $\triangle$ADE as we know that,
The sum of the angles of a triangle $= 180°$
⇒ $\angle ADE +\angle DAE + \angle AED = 180 °$
⇒ $\angle ADE +\angle DAE + 40 ° = 180 °$
$\therefore\angle ADE + \angle DAE = 140 °$
Hence, the correct answer is 140°.
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