Intersection of a cevian with the incircle

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Intersection of a cevian with the incircle

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Source: South African MO 2005 Q4

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445 posts

djb86

#1 h May 27, 2012, 9:23 AM • 6 Y

Y by megarnie, mathematicsy, jhu08, centslordm, Adventure10, Mango247

The inscribed circle of triangle $ABC$ touches the sides $BC$ , $CA$ and $AB$ at $D$ , $E$ and $F$ respectively. Let $Q$ denote the other point of intersection of $AD$ and the inscribed circle. Prove that $\text{[math]}$ extended passes through the midpoint of $AF$ if and only if $AC = BC$ .

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6874 posts

#2 h May 27, 2012, 6:14 PM • 8 Y

Y by HamstPan38825, jhu08, megarnie, centslordm, Adventure10, Mango247, Sedro, Yiyj1

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r1234

#3 h May 28, 2012, 3:41 AM • 4 Y

Y by jhu08, megarnie, centslordm, Adventure10

Let $EF\cap AD=X$ .Then $E(A,Q,X,D)=-1$ . If $\text{[math]}$ passes through the midpoint of $AF$ , then $E(A,Q,X,\infty)=-1\Longrightarrow ED\parallel AB\Longrightarrow AC=BC$ .

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algebra_star1234

2467 posts

algebra_star1234

#4 h Jun 13, 2020, 11:34 PM • 2 Y

Y by megarnie, centslordm

let $M = AF \cap EQ$ and $P = ED \cap AB$ . We know $EDQF$ is harmonic, so $-1 = (EF;QD) \stackrel E = (AF;MP)$ . If $M$ is the midpoint, then $(AF;M \infty) = -1$ , so $ED || AB$ , from which it is clear that $AC = BC$ . If $AC = BC$ , then $DE || AB$ , and we have $(AF;M\infty)=-1$ , so $M$ is the midpoint of $AF$ . Therefore, we are done.

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2786 posts

#5 h Sep 28, 2020, 5:49 AM • 1 Y

Y by centslordm

Let $M=EQ\cap AF$ ; note $AM=MF \iff (A, F; M, P_{\infty})=-1$ , but we have $(A, F; M, P_{\infty})\overset{E}=(E, F; Q, EP_{\infty}\cap \omega)$ . Thus, since $(E, F; D, Q)=-1$ , it's equivalent to $EP_{\infty}\cap \omega=D$ , or $DE||AB$ , which in turn is equivalent to $CA=CB$ .

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pad

#6 h Dec 29, 2020, 6:59 AM • 4 Y

Y by centslordm, Mango247, Mango247, Mango247

Let $\omega$ be the incricle. Define $Y\in \omega$ such that $\overline{EY}\parallel \overline{AB}$ . Then $-1=(AF;X\infty)\stackrel{E}{=}(EF;QY)$ . But $(EF;QD)=-1$ since $A=\overline{EE}\cap \overline{FF}\cap \overline{QD}$ . Hence $Y=D$ , so $\overline{ED} \parallel \overline{AB}$ . Therefore, $1=CE/CD=CA/CB$ , so $CA=CB$ . The opposite direction is very similar.

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8857 posts

#7 h Sep 5, 2021, 2:51 AM • 1 Y

Y by centslordm

$-1 = (A, F; \overline{QE} \cap \overline{AF}, P_\infty) \stackrel E= (A, \overline{EF} \cap \overline{AD}; Q, \overline{EP_\infty} \cap \overline{AD}).$ But $(A, \overline{EF} \cap \overline{AD}, Q, D)=-1$ , so $\overline{DE} \parallel \overline{AB}$ and $AC=BC$ .

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1522 posts

#8 h Sep 23, 2021, 11:07 PM • 1 Y

Y by centslordm

First, we have $(E,F;Q,D)=-1$ . Now we assume $\text{[math]}$ passes through the midpoint $M$ of $AF$ . Now if $P_\infty$ is the point at infinity along line $AB$ , we have $(A,B;F,P_\infty)=-1$ . Projecting through $E$ gives $(E,F;Q,EP_\infty \cap AD)=-1$ , which is only possible if the final point is $D$ , i.e. $ED||AB$ which implies the problem.

This post has been edited 2 times. Last edited by bever209, Sep 23, 2021, 11:08 PM

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322 posts

BVKRB-

#9 h Sep 24, 2021, 11:44 AM • 1 Y

Y by centslordm

Let $EQ \cap AF =M$
I have used the Incircle Polars Lemma (EGMO Lemma 9.27) and the Midpoints and Parallels Lemma (EGMO Lemma 9.8)
$-1 = (E,F;Q,D)\stackrel{E}{=}(A,F;M,ED \ \cap \ AB) \ \text{but} \ AM=MF \iff ED \ \cap \ AB = P_\infty \iff AC=BC \ \ \blacksquare$

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1080 posts

#10 h Dec 5, 2022, 3:56 AM • 1 Y

Y by centslordm

Let $P=\overline{QE}\cap\overline{AB}.$ Notice $-1=(Q,D;E,F)\stackrel{E}=(P,\overline{DE}\cap\overline{AB};A,F)$ so $P$ is the midpoint of $\overline{AF}$ if and only if $\overline{DE}\parallel\overline{AB}.$ But $CD=CE$ so $\overline{DE}\parallel\overline{AB}$ is equivalent to $CA=CB,$ as desired. $\square$

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4184 posts

#11 h Feb 2, 2023, 3:16 AM • 1 Y

Y by centslordm

Note that due to tangents at $A$ , $EDFQ$ is harmonic. Let $\text{[math]}$ intersect $AB$ at $K$ .

If $CA=CB$ , then $ED$ is parallel to $AB$ , so projecting through $E$ we have $-1=(EF;QD)=(AF;K\infty),$ so $K$ must be the midpoint of $AF.$

On the other hand, if $CA\neq CB$ , then $ED$ is not parallel to $AB$ . Therefore, in this case let $ED$ and $AB$ intersect at $P$ . We would then have $-1=(EF;QD)=(AF;KP),$ but since $P$ is a finite point, $K$ is not the midpoint of $AF$ in this case, so we are done.

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eibc

#12 h Aug 10, 2023, 2:37 AM • 1 Y

Y by centslordm

Let $M' = EQ \cap AF$ .. Then since $EF$ is the polar of $A$ wrt the incircle, we have $(A, EF\cap QD;Q, D) = -1$ . Taking perspectivity at $E$ gives
$-1 = (A, EF\cap QD;Q, D) \overset{E}{=} (A, F; M', DE \cap AB).$ Thus, since $(A, F; M, P\infty) = -1$ , where $M$ is the midpoint of $\overline{AF}$ and $P\infty$ is the point at infinity along line $AB$ , we have $M = M'$ iff $DE \parallel AB \iff AC = BC$ .

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5001 posts

#13 h Aug 15, 2023, 5:17 PM • 1 Y

Y by centslordm

Let $\overline{EQ} \cap \overline{AF}=T$ and $P \neq Q$ be the point on the incircle such that $\overline{PQ} \parallel \overline{AB}$ . Then $T$ is the midpoint iff $(A,F;T,P_\infty)=-1$ . We have $(A,F;T,P_\infty)\stackrel{Q}{=}(D,F;E,P)$ , so this is equivalent to $\overline{EP}$ passing through the intersection of the tangents through $D$ and $F$ , which is just $B$ . If $AC=BC$ this is clearly true by symmetry. Otherwise, note that since $DEPQ$ is cyclic, by Reim's $DEAB$ should be as well, hence by power of a point $CE\cdot CA=CD\cdot CB$ , but since $CD=CE$ this is a contradiction. $\blacksquare$

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1525 posts

#14 h Aug 30, 2023, 2:03 AM

Y by

Note that $QFED$ is harmonic. Let $M$ be the midpoint of $AF$ .
As such, $E$ , $Q$ , $M$ are collinear if and only if $(EF;QD) \overset{E}= (A,F;M,\overline{ED} \cap \overline{AB}) = -1.$ This in turn in only holds if $ED \parallel AB$ which holds when $\frac{AC}{BC} = \frac{EC}{DC} = 1.$

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Batsuh

#15 h Aug 30, 2023, 10:04 AM

Y by

First we prove that if $AC=BC$ , then $\text{[math]}$ passes through the midpoint of $AF$ .
Let $P$ be the intersection of $AD$ and $EF$ . Also let line $\text{[math]}$ intersect $AF$ at $M$ . It is clear that $QFDE$ is harmonic. Projecting through
$F$ onto $AB$ gives $(A,P;Q,D)=1$ . Thus $(EA,EP;EQ,ED)=(EA,EF;EM,ED)$ is a harmonic pencil. On the other hand $ED$ is parallel to $AF$ . This means that $M$ is the midpoint of $AF$ .
For the converse direction, assume that $M$ is the midpoint of $AF$ . Note that it's enough to show that $ED$ is parallel to $AB$ . Again, $(A,P;Q,D)=1$ and $(EA,EP;EQ,ED)=(EA,EF;EM,ED)$ is a harmonic pencil. But $M$ is the midpoint of $AF$ . Thus $ED$ is parallel to $AB$ . We're done.

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478 posts

#16 h Dec 30, 2023, 6:27 PM

Y by

Let $M=EQ\cap AF$ .

Now $-1 = (E,F;Q,D) \overset{E}{=}(A,F;M,ED\cap AB)$ .

So now, $M$ is midpoint of $AF \iff ED\parallel AB$ and we are done. :yoda:

:yoda:

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Om245

#17 h Jan 1, 2024, 7:26 PM

Y by

Let $M = \overline{AF} \cap \overline{EQ}$ As its given $MF = MA$ by POP we get

$\angle FAD = \angle AEQ \implies 2\angle BAC = 2\angle EQD = \angle DIE$ hence $\angle C = 180 - \angle A \implies CA=CB$ . $\blacksquare$

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#18 h Jan 13, 2024, 7:57 PM

Y by

Oops.

Let $M = QE \cap AF$ and $P$ as the intersection of the incircle with the line through $Q$ parallel to $AB$ . Then
$\begin{align*} MA = MF &\iff (AF;M \infty) = -1 \iff (DF;EP) = -1 \\ &\iff B, P, E \text{ collinear} \iff ABDE \text{ cyclic} \iff AC = BC. \quad \blacksquare. \end{align*}$

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cursed_tangent1434

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cursed_tangent1434

#19 h Jan 14, 2024, 3:51 AM

Y by

Quite trivial. Let $M=\overline{QE} \cap \overline{AB}$ .

We first make the following observation. Since $AE,AF$ are tangents and $A-Q-D$ we have that,
$(QD;EF)=-1$ Now, if $CA=CB$ , then $ED \parallel AB$ since $2\measuredangle CED = \measuredangle ECD = 2\measuredangle CAB$ . Thus,
$(AF;MP_\infty) \overset{E}{=}(EF;QD)=-1$ This means that $M$ is in fact the midpoint of $AF$ as desired.

If $M$ is the midpoint of $AF$ , then note that
$-1=(EF;QD) \overset{E}{=} (AF;MP)$ where $P= \overline{ED} \cap \overline{AB}$ . But, it is well known that $(AF;MP_\infty)=-1$ which implies that $P=P_\infty$ and thus, $ED \parallel AB$ which inturn implies that $\triangle CAB$ is isoceles with $CA=CB$ as desired.

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#20 h Feb 8, 2024, 10:30 AM

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$[asy] import olympiad; import cse5; defaultpen(fontsize(10pt)); usepackage("amsmath"); usepackage("amssymb"); usepackage("gensymb"); usepackage("textcomp"); size(8cm); pair MRA (pair B, pair A, pair C, real r, pen q=anglepen) { r=r/2; pair Bp=unit(B-A)*r+A; pair Cp=unit(C-A)*r+A; pair P=Bp+Cp-A; D(Bp--P--Cp,q); return A; } pointpen=black+linewidth(2); pen polyline=linewidth(pathpen)+rgb(0.6,0.2,0); pen polyfill=polyline+opacity(0.1); pen angleline=linewidth(pathpen)+rgb(0,0.4,0); pen anglefill=angleline+opacity(0.4); markscalefactor=0.01; size(12cm); pair A=dir(180+50),B=dir(-50),C=dir(90); // filldraw(A--B--C--cycle,polyfill,polyline); D(A--B--C--cycle,polyline); pair I=incenter(A,B,C); pair D=foot(I,B,C),E=foot(I,C,A),F=foot(I,A,B); pair Q=2*foot(I,A,D)-D; pair M=(A+F)/2; D(CP(I,D)); D(A--D); D(E--M,pathpen+dashed); D("A",D(A),A); D("B",D(B),B); D("C",D(C),C); D("I",D(I),dir(90)); D("D",D(D),unit(D-I)); D("E",D(E),unit(E-I)); D("F",D(F),unit(F-I)); D("Q",D(Q),dir(174)); D("M",D(M),S); [/asy]$

Let $M$ be the midpoint of $\overline{AF}$ . Now $-1 = (DQ; EF) \stackrel{E}{=} (DE \cap AB, EQ \cap AF; AF)\text{.}$ Now if $AC = BC$ then $DE \parallel AB$ so $EQ \cap AF$ must necessarily be $M$ . Conversely, if $EQ \cap AF = M$ then $DE \cap AB = \infty_{AB}$ and so $AC = BC$ .

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#21 h Feb 8, 2024, 6:23 PM

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Let $M$ be the midpoint of $AF$ .
If $AC = BC$ then it follows that $DE \parallel AB$ so we have $-1 = (P_{\infty}, M; F, A)$ . Since $QEDF$ is a harmonic quadrilateral, we have $-1 = (Q, D; E, F) \overset{E} = (\overline{EQ} \cap \overline{AB}, P_{\infty}; F, A)$ so $\overline{EQ} \cap \overline{AB} = M$ .

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Markas

#22 h Jun 16, 2024, 7:52 PM

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Let $EQ\cap AF = M$ . If we have AM = MF, then $(A, F; M, P_{\infty}) = -1$ , but also $(A, F; M, P_{\infty})\stackrel{E}{=}(E, F; Q, EP_{\infty}\cap \omega) = -1$ . Now from AF and AE tangents and A, Q, D being collinear we have that EQFD is a harmonic quadrilateral $\Rightarrow$ $(E, F; Q, D) = -1$ and from $(E, F; Q, EP_{\infty}\cap \omega) = (E, F; Q, D) = -1$ $\Rightarrow$ $EP_{\infty} \cap \omega = D$ , or $DE \parallel AB$ , which is equivalent to CA = CB, since CE = CD $\Rightarrow$ we are ready.

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bebebe

#23 h Aug 15, 2024, 1:50 PM

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We know $(E,F;Q,D)=-1$ (from tangents at $A$ ). Taking perspectivity from $E$ onto line $AB$ gives $(A,F;EQ\cap AB,ED\cap AB)=-1.$ Thus, $EQ\cap AB$ is the midpoint of $AF$ iff $ED \cap AB = P_{\infty}$ . Since $CE=CD,$ we know by similar triangles $ED \parallel AB$ iff $AC=BC$ , and we are done.

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N3bula

#24 h Dec 7, 2024, 6:20 AM

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Let $EF\cap AD$ be $P$ , we let $ED\cap AB$ be $T$ and we let $EQ\cap AF$ be $M$ , we get $-1=(A,P;Q,D)\overset{\mathrm{E}}{=}(A,F;M,T)$ as $M$ is the midpoint
of $AF$ we get that $T$ is the point at infinity so $ED$ is parallel to $AB$ so $AC=BC$ .

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#25 h Mar 30, 2025, 5:29 PM

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$-1 = (A, F; \overline{QE} \cap \overline{AF}, P_\infty) \stackrel E= (A, \overline{EF} \cap \overline{AD}; Q, \overline{EP_\infty} \cap \overline{AD}).$ But $(A, \overline{EF} \cap \overline{AD}, Q, D)=-1$ , so $\overline{DE} \parallel \overline{AB}$ and $AC=BC$ .

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