Practicing Success
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?
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\(\frac{y}{2}\) 5y 1.5y y |
y |
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 |