Question

Two lines l and M intersect at a point O and P is a point on a line n passing through the point O such that P is equidistant from L and B . Prove that N is the bisector of the angle formed by L and M.

Answer 1

Answer:

Refer to the attachment for diagram.

Given that:

Lines 'l' and 'm' intersect at O. Another line 'n' passing through O has a point 'P' which is equidistant from C and B. That, is PB = PC

According to angle bisector theorem converse,

∠ PBA = ∠ PCA = 90° [ Since they are the shortest distances from their respective points ]

To Prove: ∠ PAB = ∠ PAC

Proof:

Consider Δ PBA and Δ PCA

∠ PBA = ∠ PCA ( R ) ( 90° )

AP = AP ( H ) ( Common )

PB = PC ( S ) ( Given )

Hence by RHS congruence rule,

Δ PBA ≅ Δ PCA

By CPCT,

∠ PAC = ∠ PAB

⇒ Line 'n' is the bisector of angle A

Hence proved !!

Answer 2

HEY THERE!!!

$\huge{\bold{SOLUTION:-}}$

Given: Two lines l and M intersect at a point Of and P respectively!

And it's line passing through the at '0', Such that line P is equidistant from lines L and B.

Thus, From the above description PB = PC ( it's equidistant from line P )

Hence, Using C.P.C.T



Please refer to attachment!

Hence,

∠PAC = ∠PAB, So that it's ( OP) Bisect ∠PAC and ∠PAB.

Proved!!