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 length 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.

Steps of construction :

1. Construct a circle with centre as O and radius 4 cm.

2. Take a point A on the circle. From A make arcs of radius 6 cm and 5 cm and where they intersect the circle mark those points as B and C respectively.

(i) We know that locus of points that are equidistant from two points is the perpendicular bisector of line segment joining those points.

So from the figure,

IH is the locus of points inside the circle, that are equidistant from A and C.

Hence, the locus is the diameter of the circle which is perpendicular to the chord AC.

Proof:

Consider △GPA and △GPC.

∠PGC = ∠PGA = 90°

PG = PG (Common side)

CG = AG (as GH bisects AC)

Hence, by SAS congruence, △GPA ≅ △GPC.

Therefore, AP = PC. [By C.P.C.T.C.]

Hence, every point on the perpendicular bisector of AC is equidistant from A and C.

(ii) We know that locus of points that are equidistant from two lines is the angular bisector of the lines.

So, from the figure,

AZ is the angular bisector of angle between AB and AC.

Hence, locus is the chord of the circle bisecting ∠BAC.