In the given figure, DE is tangent to the circle at a point B. $\Delta \text{ABC}$ is constructed such that A and C are points on the circumference of the given circle, with $\text{AB} = \text{AC}$. If $\angle \text{ABD} = 65^\circ$, then what is $\angle \text{BAC}$? (Figure is not to scale)
To solve the problem, let's analyze the given conditions and apply the relevant geometric theorems:
- Given: DE is a tangent to the circle at point B, and \(\Delta \text{ABC}\) is an isosceles triangle with \(\text{AB} = \text{AC}\).
- Angle at Tangent: Since DE is a tangent, \(\angle \text{ABD} = 65^\circ\) is equal to \(\angle \text{ACB}\) because the angle between the tangent and the chord is equal to the angle in the alternate segment.
- Angles in Isosceles Triangle: In \(\Delta \text{ABC}\), since \(\text{AB} = \text{AC}\), the base angles are equal: \(\angle \text{ABC} = \angle \text{ACB}\).
- Calculation of Remaining Angle:
- Since \(\angle \text{ACB} = 65^\circ\), we have \(\angle \text{ABC} = 65^\circ\).
- Using the angle sum property of triangles: \(\angle \text{BAC} = 180^\circ - (\angle \text{ABC} + \angle \text{ACB}) = 180^\circ - (65^\circ + 65^\circ)\).
- Therefore, \(\angle \text{BAC} = 50^\circ\).
Conclusion: The measure of angle \(\angle \text{BAC}\) is 50°.
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Important Questions from Circles, Chords and Tangents
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