tangent segment to circle inside a cyclic quadrilateral

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tangent segment to circle inside a cyclic quadrilateral

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Source: Mathematical Ashes 2010

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#1 h Jul 1, 2018, 10:36 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Given a cyclic quadrilateral $ABCD$ , let the diagonals $AC$ and $BD$ meet at $E$ and the lines $AD$ and $BC$ meet at $F$ . The midpoints of $AB$ and $CD$ are $G$ and $H$ , respectively. Show that $EF$ is tangent at $E$ to the circle through the points $E, G$ and $H$ .

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dangit

#2 h Jul 4, 2018, 2:48 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

using Gauss on $AEBF$ we infer that $GH$ passes through the midpoin $T$ of segment $FE$ . The condition is equivalent to $\frac{TG}{TH}=(\frac{EG}{EH})^2$ . Since $ABCD$ is cyclic $\triangle{EAB}\sim{\triangle{EDC}}$ hence $\frac{EG}{EH}=\frac{EA}{ED}=\frac{EB}{EC}$ even further $(\frac{EG}{EH})^2=\frac{A_{EAB}}{A_{ECD}}$ . In order to determine the ratio $\frac{TG}{TH}$ we use a trick, used sometimes in the proof of Gauss's line: take $M,N,P$ the midpoints of $FB,FD,BD$ . notice that ( $M,N,T$ ), ( $N,P,H$ ) and ( $P,M,G$ ) are pairs of collinear points, so $M,N,T$ lie on the support lines of the edges of $\triangle{GHP}$ . Using Menelaus we have $\frac{TG}{TH}=\frac{MH}{MP}\times{\frac{NP}{NG}}=\frac{FB}{FD}\times{\frac{FA}{FC}}=\frac{A_{FAB}}{A_{FCD}}$ . We are left to prove that $\frac{A_{EAB}}{A_{FAB}}=\frac{A_{ECD}}{A_{FCD}}$
For this take $K\in{FE\cap{AB}}$ and $L\in{FE\cap{CD}}$ . The condition is equivalent to $\frac{FK}{KE}=\frac{FL}{LE}$ , which is true since ( $E,F;K,L$ ) $=-1$

This post has been edited 1 time. Last edited by dangit, Jul 4, 2018, 2:53 PM
Reason: typo

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#3 h Jul 13, 2018, 7:46 PM • 2 Y

Y by Adventure10, Mango247

2009 IMO Shortlist G4

Proposed by David Monk, United Kingdom

This post has been edited 1 time. Last edited by Amir Hossein, Jul 13, 2018, 10:44 PM

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