In the given figure, AD is bisector of ∠CAB and BD is bisector of ∠CBF. If the angle at C is 2y, then ∠ADB is?
Answer & explanation
Correct answer: option 4
Note the direct result :
Where AD is interior angle bisector,
and BD is exterior angle bisector
Then,
*∠ADB = \(\frac{1}{2}\)∠ACB*
Therefore,
∠ADB = \(\frac{1}{2}\) × 2y = y