Let $ABCD$ be a convex quadrilateral, $I_1$ and $I_2$ be the incenters of triangles $ABC$ and $DBC$ respectively. The line $I_1I_2$ intersects the lines $AB$ and $DC$ at points $E$ and $F$ respectively. Given that $AB$ and $CD$ intersect in $P$, and $PE=PF$, prove that the points $A$, $B$, $C$, $D$ lie on a circle.
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Tags: geometry, incenter, exterior angle, geometry proposed
Charlesww
21.10.2004 18:01
For this is the first time to post a reply in English, I'm so ashamed of my poor English. To prove that the points A, B, C, D lie on a circle, we can prove that <ABD=<ACD, the same as (<ABC - <DBC) / 2 = (<BCD - <BCA) / 2 , so we should prove that <I1BC -<I2BC = <I2CB -<I1CB , the same as <I1BI2 = <I1CI2 . To prove this, now we begin to prove that B, C, I1, I2 lie on a circle: actually, <PEF = <PFE , <FCI2 = <BCI2 , <EBI1 = <CBI1 lead to that <CBI1 + <I1I2C = <CBI1 + <FCI2 + <PFE = <EBI1 + <BCI2 + <PEF = <BCI2 + <I2I1B so <CBI1 + <I1I2C = 180 :^o: , so B, C, I1, I2 lie on a circle, so A, B, C, D lie on a circle. Q.E.D
darij grinberg
30.10.2005 14:01
(I slightly change the notations.) Problem: Let ABCD be a convex quadrilateral, and let I and J be the incenters of triangles ABC and DBC, respectively. The line IJ intersects the lines AB and DC at the points E and F, respectively. Given that the lines AB and DC intersect at P, and PE = PF, prove that the points A, B, C, D are concyclic. Since PE = PF, the triangle EPF is isosceles, and thus < PEF = (180° - < EPF) / 2. But by the exterior angle theorem applied to triangle BEI, we have < BIJ = < BEI + < EBI, hence < BIJ + < BCJ = (< BEI + < EBI) + < BCJ = (< PEF + < ABC / 2) + < BCD / 2 = ((180° - < EPF) / 2 + < ABC / 2) + < BCD / 2 = (180° - < EPF + < ABC + < BCD) / 2. But 180° - < EPF = < 180° - < BPC = < PBC + < PCB (sum of angles in triangle BPC), and hence < BIJ + < BCJ = (< PBC + < PCB + < ABC + < BCD) / 2 = ((< PBC + < ABC) + (< PCB + < BCD)) / 2 = (180° + 180°) / 2 = 180°, what entails that the quadrilateral BIJC is cyclic. Hence, < BIC = < BJC. But since I is the incenter of triangle ABC, we have < BIC = 180° - < IBC - < ICB = 180° - < ABC / 2 - < ACB / 2 = 180° - (< ABC + < ACB) / 2 = 180° - (180° - < BAC) / 2 = 180° - (90° - < BAC / 2). In other words, < BIC = 90° + < BAC / 2. Similarly, < BJC = 90° + < BDC / 2. Hence, < BIC = < BJC yields < BAC = < BDC. And thus, the points A, B, C, D lie on one circle. Darij
dikhendzab
18.04.2020 21:57
Triangle $PEF$ is isosceles, so $\angle PEF=\angle I_2I_1B-\angle EBI_1=\angle I_2I_1B-\angle I_1BC$ and $\angle PFE=\angle I_1I_2C-\angle FCI_2=\angle I_1I_2C-\angle I_2CB$. Since they are equal, $\angle I_2I_1B-\angle I_1BC=\angle I_1I_2C-\angle I_2CB$ or $\angle I_2I_1B+\angle I_2CB=\angle I_1BC+\angle I_1I_2C$. Since the sum of angles in quadrilateral is $360^\circ$ we have $\angle I_2I_1B+\angle I_2CB=180^\circ$ or quadrilateral $I_1I_2CB$ is cyclic. $\angle BI_1C=\angle BI_2C$ (same arc), so also $\angle CAB=\angle CDB$ and quadrilateral $ABCD$ is cyclic.