Question

Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of lengths 6 cm and 5 cm respectively.

(i) Construct the locus of points, inside the circle, that are equidistant from A and C. prove your construction.

(ii) Construct the locus of points, inside the circle that are equidistant from AB and AC.

Answer 1

Steps of Construction:

i) Draw a circle with radius = 4 cm.

ii) Take a point A on it.

iii) A as centre and radius 6 cm, draw an arc which intersects the circle at B.

iv) Again A as centre and radius 5 cm, draw an arc which intersects the circle at C.

v) Join AB and AC.

vi) Draw the perpendicular bisector of AC, which intersects AC at M and meets the circle at E and F.

EF is the locus of points inside the circle which are equidistant from A and C.

vii) Join AE, AF, CE and CF.

Proof :

i) In ΔCME and ΔAME

CM = AM (EF is the bisector of AC)

∠CME = ∠CMA = 90°

EM = EM (Common)

∴ By side Angle side criterion of congruence,

ΔCME ≅ ΔAME (SAS Postulate)

The corresponding parts of the congruent triangles are congruent.

⟹ CE = AE (CPCT)

Similarly, we can prove that CF = AF

Hence EF is the locus of points which are equidistant from A and C.

ii) Draw the bisector of angle A which meets the circle at N.

Therefore, Locus of points inside the circle which are equidistant from AB and AC is the perpendicular bisector of angle A.