# Use ruler and compasses only for this question.

4 views

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 f 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.

by (-3,446 points)

(i) Draw PQ, the perpendicular bisector of chord AC. PQ is the required locus, which is the diameter of the circle.

Reason : We know each point on the perpendicular bisector of AB is equidistant from A and B. Also the perpendicular bisector of a chord passes through the centre of the circle and any chord passing through the centre of the circle is its diameter.

PQ is the diameter of the circle.

(ii) Chords AB an AC intersects at M and N is a moving point such that LM = LN, where LM ⊥ AB and

LN ⊥ AC

In  right  ΔANL and ΔALB

∠ANL = ∠ABL   (90 ̊ each)

AL = AL       (Common)

NL = BL         [Given]

∴   ΔANL =  ΔALB    [R.H.S.]

Hence    ∠MAL = ∠BAL       c.p.c.t.

Thus, L lies on the bisector of  ∠BAC.

Hence  proved.