In the adjoining figure, BE and CE are bisectors of \(\angle\text{ABC}\) and \(\angle\text{ACD}\)…

In the adjoining figure, BE and CE are bisectors of \(\angle\text{ABC}\) and \(\angle\text{ACD}\) respectively. If \(\angle\text{BEC}=25^\circ\) then \(\angle\text{BAC}\) is equal to:

  1. \(65^\circ\)

  2. \(12\frac{1}{2}^\circ\)

  3. \(25\frac{1}{2}^\circ\)

  4. \(50^\circ\)

Solution

$ \angle \text{BEC} + \angle \text{EBC} = \angle \text{ECD} $ (Exterior angle property) $ \angle \text{BEC} = \angle \text{ECD} - \angle \text{ECD} $ In $ \triangle \text{ABC} $ $ \angle \text{ABC} + \angle \text{BAC} = \angle \text{ACD} $ $ \angle \text{ABC} + 2\angle \text{EBC} = 2\angle \text{ECD} $ $ \angle \text{ABC} = 2(\angle \text{ECD} - \angle \text{EBC}) $ $ \angle \text{ABC} = 2(\angle \text{BEC}) $ $ \angle \text{ABC} = 50^\circ $

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