Question

Given ∠BAC (Fig), determine the locus of a point which lies in the interior of ∠BAC and equidistant from two lines AB and AC.

Answer 1

Given: ∠BAC and an interior point P lying in the interior of ∠BAC such that PM = PN.
Construction: Join Ao and produced it to X.
Proof: In right Δs APM and APN we have
PM = PN    ...[Given]
AP = AP     ...[Common]
So, by RHS criterion of congruence

ΔAPM = ΔAPN
⇒ ∠PAM = ∠PAN  ...[∵ Corresponding parts of congruent triangle are equal]
⇒ AP is the bisector of ∠BAC
⇒ P lies on the bisector of ∠BAC
Hence, the locus of P is the bisector of ∠BAc.
Now, we shall show that every point on the bisector of ∠BAC is equidistant from AB and AC.
So, let P be a point on the bisector AX of ∠BAC and PM ⊥ AB and PN ⊥ AC. Then, we have to prove that Pm = PN.
In Δs PAM and PAN, we have
∠PAM = ∠PAN    ...[∵ AX is the bisector of ∠A]
∠PMA = ∠PNA   ...[Each equal to 90°]
and AP = AP        ...[Common]
So, by AAS criterion of congruence
ΔPAM = ΔPAN
⇒ PM = PN  ...[∵ Corresponding parts of congruent triangles are equal]
Hence, the locus of point P is the ray AX which is the bisector of ∠BAC.