easy concept,difficult provement

[photon]

At last i have proved it. i think many of you may know it,who don't,try it.
prove that ,any triangle having 2 equal angle bisector(each measured from a vertex to the opposite side) is isosceles.
[by the way,it is known as steiner-lehmus theorem.]

[photon]

none seems interested.i am giving my proof.
in a triangle $ABC$ the angle bisectors of angle $C$ and angle $B$ are CQ=BP.they intersect at $O$.add $A,O$,$AO$ intersect $BC$ at $D$.let,angle $C=2p$ and angle $B=2q$
let $p>q$
$\sin\angle p>sin\angle q$
$\frac{sin\angle p}{sin\angle q}=\frac{OB}{OC}$
$OB>OC$
IN triangle AQC,$\frac{sin\angle p}{sin\angle A}=\frac{AQ}{CQ}$........(1)
in triangle APB, $\frac{sin\angle A}{sin\angle Q}=\frac{BP}{AP}$.........(2)
multiplying 1 and 2,$\frac{sin\angle p}{sin\angle q}=\frac{AQ}{AP}$
so,$AQ>AP$.............(3)
HERE,$CQ=BP$ $OC+OQ=OB+OP$ $OQ>OP$..........(4) [as $OB>OC$]
$AB>AC$
AD bisects angle $CAB$
$\frac{AB}{AC}=\frac{BD}{CD}$ so $\frac{BD}{CD}>1$..........(5)
in triangle COB,$\frac{sin\angle BOD}{sin \angle COD}=\frac{OC}{OB}\frac{BD}{CD}$
in triangle APQ,$AQ>AP$ SO, $\angle APQ>\angle AQP$ ........(6)
in triangle OPQ,$OQ>OP$ SO, $\angle OPQ>\angle OQP$........(7)
ADDING 6 AND 7,$\angle APO>\angle AQO$
in triangle AOP and AOQ,$\angle APO>\angle AQO$
$\angle PAO=\angle QAO$ [as AO bisect angle A]
$\angle AOQ>\angle AOP$ or, $\angle COD>\angle BOD$
then,$\frac{sin\angle BOD}{sin \angle COD}<1$
$\frac{OC}{OB}\frac{BD}{CD}<1$ [5 no. equation]
as $OB>OC$ so, $CD>BD$.but we have showed $\frac{BD}{CD}>1$ $CD<BD$ contradiction(at last! )
so,p=q.the triangle is a isosceles triangle.