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Pirkuliyev Rovsen

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Pirkuliyev Rovsen

#1 h Dec 1, 2013, 5:07 PM • 2 Y

Y by Adventure10, Mango247

Let $CC_1$ be the angle bisector and $J$ the incenter of triangle $ABC$ , ${\angle}A=60^\circ$ , ${\angle}B=80^\circ$ .Prove that $P_{{\bigtriangleup}BJC_1}=BC$ .

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SayanRoychowdhuri

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SayanRoychowdhuri

#2 h Dec 1, 2013, 9:50 PM • 2 Y

Y by Pirkuliyev Rovsen, Adventure10

Pure Geometric Proof: let angular bisector of $\angle B$ meet $AC$ at $K$ clearly as $\angle A=60^0$ so $AC_1JK$ is cyclic and $C_1J=JK$ now we extend $BK$ upto $D$ such that $BD=BC$ Now draw line $KL$ through $K$ And parallel to $BC$ such that $BK=KL$ now clearly $BK=CK$ , hence we get $BK=CK=KL$ also we have $\angle KDC=\angle KLC=70^0$ , hence we get $KDLC$ is cyclic, so , $\angle DLK=30^0$ and readily we have $\angle DKL=40^0$ ,now compairing $\triangle KDL$ and $\triangle BC_1K$ we have $BK=KL$ and also $\angle C_1KB=\angle DLK=30^0$ and other angles are also equal, hence they are congruent, so, $BC_1=KD$ now $BJ+JC_1+BC_1=BJ+JK+KD=BD=BC$ , so, we are done

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