# A question on Trigonometry (bisector)

If two bisector of a triangular is equal, then it is Isosceles triangular.

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What do you men by bisector of a triangular? –  Abhijit Mar 25 '13 at 5:51
"If two bisectors of a triangle have equal length, then the triangle is isosceles?"? –  alex.jordan Mar 25 '13 at 7:21

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 $BE<CL$. Hence, $BE<CL<CF$

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OE and OD are not inradii, they are segments on the bisectors. –  zyx Mar 25 '13 at 21:25