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Question
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
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Solution
Steps of construction:
- Draw a ray BC.
- Construct a ray RA making an angle of 75° with BC. Therefore, ABC = ABC = 75°
- Draw the angle bisector BP of ∠ABC.
BP is the required locus. - Take any point D on BP.
- From D, draw DE ⊥ AB and DF ⊥ BC.
Since D lies on the angle bisector BP of ∠ABC.
D is equidistant from AB and BC.
Hence, DE = DF
Similarly, any point on BP is equidistant from AB and BC.
Therefore, BP is the locus of all points which are equidistant from AB and BC.
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