The external bisector of angle A of A triangle ABC meets BC produced at L. And the internal bisector of angle B meets CA at M. If LM meets AB at R. Prove that CR bisects angle C?

Question

Answer

Check your question

Answer

Given:

Triangle ABC

External bisector of angle A meets BC produced at L

Internal bisector of angle B meets CA at M

LM meets AB at R

To prove: CR bisects angle C

Proof:

Lemma 1: The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the adjacent sides.

Proof of Lemma 1: Let ABC be a triangle and AD be its external bisector of angle A. Then, by angle bisector theorem, we have:

BD/DC = AB/AC

By exterior angle theorem, we have:

∠BAC = ∠ABD + ∠ADC

But, ∠ABD = ∠ADC (as AD is an external bisector)

So, ∠BAC = 2∠ABD

Therefore, in triangle ABD, we have:

sin(∠ABD)/BD = sin(∠BAC/2)/AB

Similarly, in triangle ADC, we have:

sin(∠ADC)/DC = sin(∠BAC/2)/AC

Dividing the two equations, we get:

BD/DC = AB/AC

which proves the lemma.

Lemma 2: The internal bisectors of two angles of a triangle intersect at a point which is equidistant from the sides of the third angle.

Proof of Lemma 2: Let ABC be a triangle and AD and BE be its internal bisectors of angles A and B respectively. Then, by angle bisector theorem, we have:

BD/DC = AB/AC ...(1)

and

AE/EC = AB/BC ...(2)

Multiplying (1) and (2), we get:

(BD/DC)*(AE/EC) = 1

By converse of angle bisector theorem, we have:

AD and BE intersect at a point I which is equidistant from sides AC and BC.

Proof of the main result: Since CR is the third transversal of the lines AL and BM, we have:

(BL/LC) * (CA/AM) * (MR/RB) = 1

By Lemma 1, we have:

BL/LC = AB/AC

By Lemma 2, we have:

CA/AM = CB/BM

Therefore, we have:

(AB/AC) * (CB/BM) * (MR/RB) = 1

which implies:

MR/RB = AC/AB

Similar Class 10 Doubts

In any triangle ABC, if angle B > angle C then prove that the internal bisector BE of angle B is shorter than the internal bisector CF of angle C.?

The side BC of a triangle ABC is bisected at D;O is any point in AD.BO and CO produced meet AC and AB in E and F respectively and AD is produced to X so that D is the mid point of OX. Prove that AO:AX = AF:AB and show that FE || BC?

A single question to geniuses . #Mathematics. A triangle ABC is circumscribed inside a circle with centre O . AB=AC ,

D is the mid point of side BC of a triangle ABC .AD is bisected at the point E and BE produced cuts AC at the point X . Probe that BE:EX=3:1?

If an angle bisector of a triangle bisects it's opposite side, then the triangle is

Top Courses for Class 10View all

Science Class 10

Social Studies (SST) Class 10

Mathematics (Maths) Class 10

English Grammar Advanced

English Grammar Basic

Top Courses for Class 10

Top Courses for Class 10

Suggested Free Tests

Related Content

About Class 10

Class 10 Study Material

Class 10 All Subjects

NCERT Textbook Solutions

Other Classes

Important Links

Company