If two bisectors of a triangle are equal, then the triangle is isosceles.

Question

If two bisectors of a triangle are equal, then the triangle is isosceles.

Answer 1

This is the Steiner-Lehmus problem.

The most straightforward proofs use algebra or trigonometry to get a formula for the length of the bisector as a function of the sides of the triangle. Geometric proofs are more complicated and require the use of inequalities. If there is no explicit use of inequalities, the proof must rely on some assumption about the position of points in the diagram (such as whether they are inside the triangle or not) and this has led to some false proofs in which the assumption is wrong.

Answer 2

This is a famous Steiner-Lehmus problem

1.

**This proof here is not right, this is the common mistake people do ** Lets say $O$ is the concurring point of the bisectors.

∴ $O$ will be the Incentre and OE,OD will be In-radii. Thus equal.

∴ $BO = OC$ ( Subtracting $OE,OD$ from $CE$ & $BD$ )

∴ $\angle DBC = \angle ECB$

∴ $\angle ABC = \angle ACB $

∴$AB = AC$

But here, $OE$ and $OD$ will not be inradii. They are not even part of the in-circle(Unless you prove it)

2.

If $AB=AC$, what can we say about the relationship between $BD$ and $CE$.

Observe that $\angle ABC =\angle ACB \implies \angle DBC = \angle ECB$, Thus $BD=CE$

3.

There's theorem which says if $\angle ABC > \angle ACB$, then angle bisector of $\angle ABC$ < angle bisector of $\angle ACB$.

Proof: Since $\angle ABC > \angle ACB$, $\angle ABD> \angle ACE$. Let X be a point on line segment $AD$ such that $\angle XBD=\angle ACE$.

Now, $BD$ and $CE$ meet at $O$(Incentre). Let $BX$ meet $CE$ at $L$.

By construction $\triangle XBD$ and $\triangle XCL$ are similar.

Therefore, $\dfrac{BD}{CL}=\dfrac{BX}{CX}$

In $\triangle XBC$, we have $\angle XBC= \dfrac{\angle B}{2}+\dfrac{\angle C}{2}>+\dfrac{\angle C}{2}+\dfrac{\angle C}{2}=\angle XCB$

And hence $XC>BX$.

Therefore, we have $1<\dfrac{BX}{XC}=\dfrac{BD}{CL}$ or $BD<CL$. Hence, $BD<CL<CE$