Given an equilateral $\Delta ABC$, in which ${{A} _ {1}}, {{B} _ {1}}, {{C} _ {1}}$ are the midpoint of the sides $ BC, \, \, AC, \, \, AB$ respectively. The line $l$ passes through the vertex $A$, we denote by $P, Q$ the projection of the points $B, C$ on the line $l$, respectively (the line $ l $ and the point $Q, \, \, A, \, \, P$ are located as shown in fig.). Denote by $T $ the intersection point of the lines ${{B} _ {1}} P$ and ${{C} _ {1}} Q$. Prove that the line ${{A} _ {1}} T$ is perpendicular to the line $l$. (Serdyuk Nazar)
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Tags: geometry, perpendicular, Equilateral
06.08.2020 00:35
[asy][asy] unitsize(2cm); pair A=dir(90); pair B=dir(210); pair C=dir(330); dot(A); dot(B); dot(C); draw(A--B--C--cycle); pair X=(1,0.2); draw((A-2*X)--(A+2*X)); pair P=foot(B,A,(A-X)); draw(P--B); pair Q=foot(C,A,(A-X)); draw(Q--C); draw(A/2+B/2--Q); draw(A/2+C/2--P); pair T=extension(A/2+B/2, Q, A/2+C/2, P); dot(T); draw(B/2+C/2--extension(B/2+C/2,T,P,Q)); label("$A$",A,NNW); label("$B$",B,SW); label("$C$",C,SE); label("$P$",P,NNW); label("$Q$",Q,NNW); label("$A_1$", B/2+C/2, S); label("$B_1$", A/2+C/2, NNE); label("$C_1$", B/2+A/2, WNW); label("$T$", T, SE); draw(circumcircle(A,Q,C)); [/asy][/asy] We show that $TP=TQ$ and $A_1P=A_1Q$. Since $\angle AQC$, $\angle AA_1C$, and $\angle AC_1C$ are all $90^\circ$, $AC_1A_1CQ$ are is a cyclic pentagon. Hence $\angle C_1QC=\angle C_1AC=60^\circ$. Hence $\angle AQC_1=30^\circ$. Similarly, $B_1PA=30^\circ$, so $\triangle PTQ$ is isosceles. Also, $\angle AQA_1=\angle ACA_1=60^\circ$ and similarly $\angle APA_1=60^\circ$, so $PA_1Q$ is isosceles. Hence both $T$ and $A_1$ lie on the perpendicular bisector of $PQ$ whence $A_1T\perp PQ$.