(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.