# Geometry Problem on triangle?

Using pure geometry, prove that if two angle bisectors of a triangle are of equal length, it is an isosceles triangle. Do not use coordinate geometry or vectors or trigonometry.

I had found the proof when I was in college, but after that I am not able to recollect it despite having spend hours on it so many times. I remember having started with the assumpsion "let the sides opposite to the angles bisectors be unequal" and then proving that it is not possible.

Neel

That theorem is of no use here. What I have asked to prove is the converse of the theorem. It has to be proved independenly and not assuming that converse is true.

prabh_rishab

OB = OC is your assumpsion and not given. It is to be proved before acceptance.

To clarify the problem, in triangle ABC, BE and CD are angle bisectors of equal lengths, where E is on AC and D is on AB.

I shall keep this problem open as long as I can.

### 9 Answers

- DukeLv 71 decade agoBest Answer
Hello, Mr. Daftary, I am glad to a chance to answer this question of Yours.

Scythian1950 is right - this is the (infamous) Steiner-Lehmus Theorem. Here is the proof from H.S.M.Coxeter's excellent book 'Geometry Revisited', I'll reproduce it almost literally.

Lemma 1.511. If 2 chords in a circle subtend different acute angles with vertexes on the circumference, then the greatest angle corresponds to the longest chord.

Proof: Equal chords subtend equal central angles, hence equal angles with vertexes on the circumference (as halves of the former). If 2 chords are non-equal , then the shorter, being farther from the center, subtends a smaller central angle, hence a smaller acute angle with a vertex on the circumference.

Lemma 1.512. In a triangle with 2 different angles the smaller angle has a longer bisector.

Proof: Let ABC is a triangle with angle(B) = β < γ = angle(C) and BE and CD are angle bisectors. Let E' is a point on BE, such that

angle(E'CD) = β/2 = angle(E'BD).

The above means the points B, D, E'. C are con-cyclic. Since

β < (β + γ)/2 < (α + β + γ)/2, α = angle(A) as usual,

angle(CBD) < angle(E'CB) < π/2 and according 1.511

| CD | < | BE' |, hence | BE | > | BE' | > | CD |.

Now the Steiner-Lehmus Theorem follows directly from 1.512 since

β ≠ γ implies | BE | ≠ | CD |

P.S.According Coxeter, this is one of the simplest proofs of many published in books and magazines.

Source(s): 'Geometry Revisited', H.S.M.Coxeter, 1967 - 1 decade ago
I remember that there is a thorem that if the two sides of a traingle are equal then the angle opposite to them are also same. Please see the thorem.

So the bisectors make a isoscele traingle with the side in between the angles were getting bisected.I will later write you the proof of the theorm.

As the each half of the angles are equal, then the full angles are equal.

here you found the two angles are equal.

As per definition: in a traingle if the two angles are equal, then that traingle is called a isoscele traingle.

- PramodLv 71 decade ago
Let us suppose that ABC is the given triangle. Angles B and C are bisected to intersect at D.

We know that D is circum-center of the Triangle ABC. Hence BD = CD

From Triangle BCD we have Angle DBC = Angle DCB = x (say) ---- (since BD = CD)

But BD bisects Angle ABC . Therefore Angle ABD too = x.

Therefore Angle ABC = 2x . In the same way we can prove that Angle ACB = 2x

Thus Angle ABC = Angle ACB = 2x each.

Hence AB = AC. …………………… Proved.

P.K.Tandon

Edit : Now I think this is sufficient.

…………………………………………………………………………..

- 1 decade ago
Hello sir!

Please if you can make an effort to draw the diagram as you read,it would be easy.

As given two bisectors are equal.

1.consider base BC and draw bisector BE which cuts AC at E.

2.consider another bisector DC, bisecting angle C and which cuts AB at D

Given: BE=DC

Therefore angle EBC = angle DCB

hence BD=EC {opposite sides to equal angles are equal}

now,

(AB/DB)=(AC/EC) {Join DE and it comes out to be parallel to base BC.This is the converse of Basic Proportionality Theorem}

Thus

(AB/AC)=(BD/EC)=1

as BD=EC {proved earlier}

thus AB=AC.

Please correct me if I am wrong

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- 1 decade ago
GIVEN

ABC is triangle with BC as base and O is point of intersection of angle bisectors from angle A , angle B & angle C. and OB = OC

TO PROVE

AB = AC

PROOF

in triangle AOB and triangle AOC

angle BAO = angle CAO-------[AO is bisector of angle A]

AO = AO-----------------------------[common]

OB = OC-----------------------------[given]

therefore triangle AOB congruent to triangle AOC by angle-side-side theorem

so AB = AC [corresponding parts of congruent triangles]

- Scythian1950Lv 71 decade ago
This is a beast problem. I proved it once before a long time ago, but I have to remember how I did it. Give me some time.

Edit: This is the infamous Lehmus-Steiner problem. The proof I had a long time ago is very similiar to the one offered in the "Ask Dr Math" forum, see link. But I will continue to try to remember exactly how I did it myself.

- VickyLv 51 decade ago
in triangle ABC seg BD and seg CD are the bisectors of the angles <B and <C respectively . BD=CD

We have to prove that AB=AC

Assume that AB is not equal to AC

<ABC is not equal to <ACB

1/2<ABC is not equal to 1/2<ACB

<DBC is not equal to <DCB

BD is not equal to CD

But it is not possible since BD=CD is given

AB must be equal to AC i.e. triangle is an isosceles triangle

- Rita the dogLv 71 decade ago
Sorry, I have no desire to be frustrated this morning. However, you may find the following link of interest: