A point $P$ lies on the internal angle bisector of $\angle BAC$ of a triangle $\triangle ABC$. Point $D$ is the midpoint of $BC$ and $PD$ meets the external angle bisector of $\angle BAC$ at point $E$. If $F$ is the point such that $PAEF$ is a rectangle then prove that $PF$ bisects $\angle BFC$ internally or externally.
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Tags: geometry, rectangle, geometric transformation, reflection, circumcircle, angle bisector, geometry unsolved
skytin
22.10.2010 19:09
Let W is midpoint of arc BC of (ABC) . let WD intersect FP at point K , let AW intersect BC at point L , let KL intersect AF at point X , easy to see that angle XAL = APE = LKD , so XAKW is cyclic , so KL*LX = AL*LW = BL*LC , so BKCX is cyclic . easy to see that angle PFA = DLK , and K is midpoint of arc BC of (BKC) , so KB is tangent to (XBL) , so angle KBX = KLC = AFK , so F is on (BKC) , K is midpoint of arc BC of (BKC) , so PF bisects angle BFC internally or externally. done
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Vo Duc Dien
31.12.2011 19:36
This is also problem 4 of the British Mathematical Olympiad 1987. The original British problem stated that "The triangle ABC has orthocenter H. The feet of the perpendicu-lars from H to the internal and external bisectors of angle BAC (which is not a right angle) are P and Q. Prove that PQ passes through the middle point of BC." BTW. I solved it years ago by employing the fact that the distance from the centroid of a triangle to the midpoint of the side is 1/2 that to the vertex.
RSM
01.01.2012 11:23
Reflect $ A,F $ on $ D $ to get $ A',F' $. Construct a quadrilateral $ IJKL $ such that midpoints of $ IJ,JK,KL,LI,IK,JL $ are $ A',C,A,B,F,F' $. Now note that bisector of the angle between $ IK $ and $ JL $ is the same as that of the angle between $ IL $ and $ JK $. So $ IJKL $ is cyclic. So bisector of the angle between $ IJ $ and $ KL $ is parallel with them too. So done. [asy][asy] import graph; size(280); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; draw((-1.62,(+4.15-5.1*-1.62)/-1.78)--(14,(+4.15-5.1*14)/-1.78)); draw((-1.62,(+2.35-0*-1.62)/6.52)--(14,(+2.35-0*14)/6.52)); draw((-1.62,(-39.75+5.1*-1.62)/-4.74)--(14,(-39.75+5.1*14)/-4.74)); draw((-1.62,(+6.5-5.1*-1.62)/4.74)--(14,(+6.5-5.1*14)/4.74)); draw((-1.62,(-37.41+5.1*-1.62)/1.78)--(14,(-37.41+5.1*14)/1.78)); draw((-1.62,(+6.85-3.72*-1.62)/-0.6)--(14,(+6.85-3.72*14)/-0.6)); draw((-1.62,(-4.79-0.21*-1.62)/-0.98)--(14,(-4.79-0.21*14)/-0.98)); draw((-1.62,(+3.78-0.98*-1.62)/0.21)--(14,(+3.78-0.98*14)/0.21)); draw((-1.62,(+19.73-3.16*-1.62)/2.04)--(14,(+19.73-3.16*14)/2.04)); draw((-1.62,(+8.32-3.16*-1.62)/2.04)--(14,(+8.32-3.16*14)/2.04)); draw((-1.62,(+23.93-3.72*-1.62)/-0.6)--(14,(+23.93-3.72*14)/-0.6)); draw((-1.62,(+6.75-1.23*-1.62)/4.4)--(14,(+6.75-1.23*14)/4.4)); draw((-1.62,(+21.95-5.1*-1.62)/1.48)--(14,(+21.95-5.1*14)/1.48)); draw((-1.62,(-0.2+0.61*-1.62)/1.06)--(14,(-0.2+0.61*14)/1.06)); draw((-1.62,(+2.54-0.61*-1.62)/5.46)--(14,(+2.54-0.61*14)/5.46)); draw((-1.62,(-4.2+0.61*-1.62)/1.06)--(14,(-4.2+0.61*14)/1.06)); draw((-1.62,(-6.55+0.61*-1.62)/-5.46)--(14,(-6.55+0.61*14)/-5.46)); draw((-1.62,(+3.28-3.72*-1.62)/-0.6)--(14,(+3.28-3.72*14)/-0.6),red); draw((-1.62,(+4.12+0.61*-1.62)/1.06)--(14,(+4.12+0.61*14)/1.06),red); draw((-1.62,(+24.34-3.16*-1.62)/2.04)--(14,(+24.34-3.16*14)/2.04),red); draw((-1.62,(-22.39-0.61*-1.62)/5.46)--(14,(-22.39-0.61*14)/5.46),red); draw(circle((6.16,-0.48),6.97),magenta); dot((2.72,5.46),ds); label("$A$", (2.83,5.65),NE*lsf); dot((0.94,0.36),ds); label("$B$", (1.05,0.56),NE*lsf); dot((7.46,0.36),ds); label("$C$", (7.59,0.56),NE*lsf); dot((4.2,0.36),ds); label("$D$", (4.34,0.56),NE*lsf); dot((5.68,-4.74),ds); label("$A'$", (5.81,-4.54),NE*lsf); dot((2,0.97),ds); label("$F$", (2.11,1.15),NE*lsf); dot((6.4,-0.25),ds); label("$F'$", (6.53,-0.07),NE*lsf); dot((1.66,4.85),ds); label("$L$", (1.77,5.03),NE*lsf); dot((3.78,6.07),ds); label("$K$", (3.9,6.25),NE*lsf); dot((11.14,-5.35),ds); label("$J$", (11.28,-5.17),NE*lsf); dot((0.22,-4.13),ds); label("$I$", (0.33,-3.95),NE*lsf); clip((-1.62,-7.86)--(-1.62,8.03)--(14,8.03)--(14,-7.86)--cycle);[/asy][/asy]
polya78
17.03.2013 23:47
Draw the circumcircle of $\triangle ABC$ and let $R,S$ be the midpoints of the arc $BC$. Then $AR, AS$ are the external and internal bisectors of $\angle BAC$. Let $X$ be on $RS$ such that $PX \parallel AR$. Let $w$ the circumcircle of $BCX$ and $Y$ be the antipode of $X$ on $w$. Draw a line parallel to $AS$ through $Y$ and let $E',F'$ be its intersections with $AR, PX$. Then $XY$ is a diameter of $w$, so $F'$ lies on $w$ and clearly $F'Y$ bisects $\angle BF'C$. If we can show that $E',P,D$ are collinear, then $E=E',F=F'$ and we are done. This is true if $RE'/PX = RD/XD$, or if $RD/XD =RY/XS$, or $(RD)(XS)=(RY)(XD)$, or $(RD)(XD+DS)=(RD+DY)(XD)$, or $(RD)(DS)=(DY)(XD)$, which is clearly true since each side is equal to $(BD)(DC)$.
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india postal 2010.pdf (417kb)