Mathematics
Using ruler and compasses only, construct an isosceles ΔABC in which BC = 5 cm, AB = AC and ∠BAC = 90°. Locate the point P such that :
(i) P is equidistant from BC and AC.
(ii) P is equidistant from B and C.
Locus
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Answer
Steps of construction :
Make BC = 5 cm as base.
Create a semicircle with BC as diameter.
Make right bisector of BC and mark it's intersection and BC as D and the semicircle as A. Join AB and AC.
Make angle bisector of ∠ACB. From graph, CE is the angle bisector.
Mark point P where angle bisector ∠ACB i.e CE meets perpendicular bisector of BC i.e. AD.
(i) We know that locus of points equidistant from two lines is the angle bisector of angle between them.
From figure,
CE is the angular bisector of ∠ACB, hence it will be equidistant from AC and BC.
(ii) We know that locus of points equidistant from two points is the perpendicular bisector of line segment joining them.
From figure,
AD is the perpendicular bisector of BC, hence it will be equidistant from B and C.
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