MB = MC wanted, projections on angle bisector, midpoint

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MB = MC wanted, projections on angle bisector, midpoint

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Source: 2002 Belarusian MO D 8.3

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#1 h Dec 4, 2020, 6:16 PM

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Points $B_1$ and $C_1$ are marked on the bisector of $\angle A$ of the triangle $ABC$ so that $BB_1 \perp AB, CC_1 \perp AC$ . Let $M$ be the midpoint of $B_1C_1$ . Prove that $MB = MC$ .

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#2 h Dec 4, 2020, 7:09 PM

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This problem is just as easy as proving Fermat's Last Theorem, but having a too short margin.

Let $D=BB_1\cap CC_1$ , let $M$ be intersection of the bisector of $\angle A$ and $(ABC)$ . Notice that $MC=MB$ , because $\angle MBC=\angle MAC=\angle BAM=\angle BCM$ .
I claim that $M$ is the midpoint of $B_1C_1$ . We have that $\angle DMA=\angle DCA=90^\circ$ .
Also, $\angle DB_1C_1=90^\circ-\angle BAM=90^\circ-\angle MAC=\angle AC_1C=\angle DC_1B_1\implies DB_1=DC_1.$ Hence, we are done.

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#3 h Dec 5, 2020, 4:34 AM

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As in the above solution, let $D=BB_1\cap CC_1$ .
Since $\angle ABD = \angle ACD = 90$ , $D$ lies on the circumcircle $\omega$ of $\triangle ABC$ and $AD$ is a diameter.
Let $M'=B_1C_1\cap \omega$ , so $M'$ is the midpoint of arc $\widehat{BC}$ of $\omega$ . I will show $M'=M$ .
Let the parallel through $D$ to $AM'$ meet $\omega$ at $N$ . Since $AD$ is a diameter, $MN'$ is also a diameter.
Projecting through $D$ onto $\omega$ ,
$(B_1,C_1;M',{B_1C_1}_{\infty})=(B,C;M',N)$ .
But $BCM'N$ is a harmonic quadrilateral (since it is a kite inscribed in a circle), so $(B,C;M',N)=1$ .
Thus, $(B_1,C_1;M',{B_1C_1}_{\infty})=\frac{B_1M'}{C_1M'}=1$ , so $B_1M'=C_1M'$ and thus $M'=M$ is the midpoint of $B_1C_1$ .
Then $MB=MC$ as $M$ is the midpoint of arc $\widehat{BC}$ of $\omega$ .

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