Mathematics
By using ruler and compasses only, construct an isosceles triangle ABC in which BC = 5 cm, AB = AC and ∠BAC = 90°. Locate the point P such that
(i) P is equidistant from the sides BC and AC.
(ii) P is equidistant from the points 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 as D.
Make angle bisector of ∠ACB. From graph, CE is the abgle bisector.
Mark point P where angle bisector ∠ACB i.e CE meets perpendicular bisector of BC i.e. AD.
Draw perpendicular from point P and it meets semicircle at point A.
(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.
Both AD and CE meet at point P, so P is equidistant from BC and CA and also from B and C.
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