Question

In the figure a is a point equidistant from to linesl1 and l2 intersecting at point P.show that AP bisects the angle between l1 and l2.

Answer 1

Solution:-
Given : L 1 and L 2 are intersecting lines, intersect at P.
A is a point such that perpendicular from P to L 1 and L 2.
Therefore AM = AN
So, AP bisects ∠ MPN
Proof :
In the triangles AMP and ANP,
AM = AN    (Since 'A' is equidistant from L 1 and L 2)
∠ AMP = ∠ ANP = 90°
AP is common.
Therefore by right angle hypotenuse side congruency criterion,
Δ AMP ≡ Δ ANP
Thus by CPCT : ∠ MPA = ∠ NPA
Thus AP bisects the angle between L 1 and L 2
Hence proved.

Answer 2

See diagram.

Given AB = AC  perpendicular distances from A to L1 and L2.

ΔAPB and ΔAPC are right angle triangles. Applying Pythagoras theorem, 
    PB² = AP² - AB²
    PC² = AP² - AC²  = AP² - AB² = PB²

ΔAPB and ΔAPC are congruent, as all corresponding sides are equal.
Hence, ∠APB = ∠APC
=>  AP is the angular bisector of ∠BPC