Show that $ca = cb$.

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Fermat -Euler

444 posts

Fermat -Euler

#1 h Nov 2, 2005, 2:05 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle, and let the angle bisectors of its angles $CAB$ and $ABC$ meet the sides $BC$ and $CA$ at the points $D$ and $F$ , respectively. The lines $AD$ and $BF$ meet the line through the point $C$ parallel to $AB$ at the points $E$ and $G$ respectively, and we have $FG = DE$ . Prove that $CA = CB$ .

Original formulation:

Let $ABC$ be a triangle and $L$ the line through $C$ parallel to the side $AB.$ Let the internal bisector of the angle at $A$ meet the side $BC$ at $D$ and the line $L$ at $E$ and let the internal bisector of the angle at $B$ meet the side $AC$ at $F$ and the line $L$ at $G.$ If $GF = DE,$ prove that $AC = BC.$

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grobber

7849 posts

grobber

#2 h Nov 23, 2005, 8:21 PM • 2 Y

Y by Adventure10, Mango247

Fermat -Euler wrote:

Let $ABC$ be a triangle, and let the angle bisectors of its angles $CAB$ and $ABC$ meet the sides $BC$ and $CA$ at the points $D$ and $F$ , respectively. The lines $AD$ and $BF$ meet the line through the point $C$ parallel to $AB$ at the points $E$ and $G$ respectively, and we have $FG = DE$ .
Prove that $CA = CB$ .

Let a=BC, b=CA, c=AB.
We have BF/AD=(BF/FG)/(AD/DE)=(c/a)/(c/b)=b/a, but it's easy to see from the formulae of the type la=2bc*cos(A/2)/(b+c) that the order of the bisectors of a triangle is inverse to that of the sides (if a>=b>=c then la<=lb<=lc). If b>a then BF/AD=b/a>1, which is false, so b<=a => contradiction. We do the same for b<a, so the only possible situation is a=b, Q.E.D.

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jgnr

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jgnr

#3 h Dec 12, 2009, 11:07 AM • 2 Y

Y by Adventure10, Mango247

another solution

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dgreenb801

1896 posts

dgreenb801

#4 h Dec 12, 2009, 3:14 PM • 2 Y

Y by Adventure10, Mango247

Let $I$ be the incenter, and let $CI$ intersect $AB$ at $P$ . Since $GE \parallel AB$ , we have $\triangle ABD \sim \triangle DEC$ and $\triangle ABF \sim \triangle FCG$ . Thus,
$\frac {BD}{DC} = \frac {AD}{DE}$ and $\frac {AF}{FC} = \frac {BF}{FG}$ . Since $DE = FG$ , dividing the two we have
$\frac {BD}{DC} \cdot \frac {CF}{FA} = \frac {AD}{BF}$ . But by Ceva, $\frac {BD}{DC} \cdot \frac {CF}{FA} = \frac {PB}{PA} = \frac {BC}{CA}$ (angle bisector theorem). So $\frac {BC}{CA} = \frac {AD}{BF}$ . But if $BC \neq CA$ , then this would mean that lengths of the angle bisectors are in the same order as the lengths of the sides they intersect, a contradiction by the generalization of the Steiner-Lehmus Theorem in the link http://www.cems.uvm.edu/~cooke/sl.pdf. Thus, $BC = CA$ .

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sayantanchakraborty

505 posts

sayantanchakraborty

#5 h Jun 5, 2014, 11:19 AM • 1 Y

Y by Adventure10

Never seen such an easy geometry in the ISLs atleast.

Wlog let $a \ge b$ .Note that $CF=\frac{ab}{a+c},\angle{GCA}=A,\angle{FGC}=\frac{B}{2}$

So application of sine rule in $\triangle{CFG}$ yeids

$\frac{GF}{sinA}=\frac{ab}{(a+c)sin\frac{B}{2}}$

Similarly sine rue in $\triangle{CDE}$ yeilds

$\frac{ED}{sinB}=\frac{ab}{b+c}$

Dividing the two relations we get

$\frac{(a+c)b}{(b+c)a}=\frac{sin\frac{A}{2}}{sin\frac{B}{2}} \ge 1$

by our assumption.

This gives

$\frac{(a+c)b}{(b+c)a} \ge 1$

$\Rightarrow ab+bc \ge ab+ac \Rightarrow b \ge a$

So we obtain $b=a$

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WolfusA

1900 posts

WolfusA

#6 h Jul 22, 2017, 6:18 AM • 1 Y

Y by Adventure10

OK, maybe some logical order. Everyone up, who made assumption that $a\geq b$ and didn't check the case $a<b$ has only a half of solution. By the way you can't have an assumption that $a=b$ , so you can't make the case $a\geq b$ , because you have to prove, that $a=b$ . How can the proof be correct, when you prove something supposing it. Correct solution is when you show, that both inequalities $a>b$ and $a<b$ are contradiction, so there must be $a=b$ .

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adamov1

355 posts

adamov1

#7 h Jul 22, 2017, 6:24 AM • 1 Y

Y by Adventure10

WolfusA wrote:

OK, maybe some logical order. Everyone up, who made assumption that $a\geq b$ and didn't check the case $a<b$ has only a half of solution. By the way you can't have an assumption that $a=b$ , so you can't make the case $a\geq b$ , because you have to prove, that $a=b$ . How can the proof be correct, when you prove something supposing it. Correct solution is when you show, that both inequalities $a>b$ and $a<b$ are contradiction, so there must be $a=b$ .

Your first complaint is incorrect because you can make the assumption $a\ge b$ because the two cases are exactly symmetric, hence the "Wlog" (just flip your diagram). In addition, your second complaint is incorrect because supposing $a\ge b$ is not the same thing as $a=b$ . Obviously at least one of $a\ge b, a\le b$ is true, and since they are symmetric you may assume one of them. You should learn some logic yourself before criticizing others.

This post has been edited 1 time. Last edited by adamov1, Jul 22, 2017, 6:26 AM

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WolfusA

1900 posts

WolfusA

#8 h Jul 22, 2017, 6:46 AM • 2 Y

Y by Adventure10, Mango247

I don't mean that WLOG part is incorrect.
So if you like so this equality in assumption check yourself that this $p \vee q \implies p$ is tautology. And put here $p: a=b$ and $q: a>b$ . Anyway the second case (after WLOG) is $a<b$ (not $b\geq a)$ .

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adamov1

355 posts

adamov1

#9 h Jul 22, 2017, 6:54 AM • 2 Y

Y by Adventure10, Mango247

WolfusA wrote:

I don't mean that WLOG part is incorrect.
So if you like so this equality in assumption check yourself that this $p \vee q \implies p$ is tautology. And put here $p: a=b$ and $q: a>b$ . Anyway the second case (after WLOG) is $a<b$ (not $b\geq a)$ .

I'm not sure exactly what you're saying (I'm pretty sure the first thing you said isn't a sentence), but $p \vee q \implies p$ is not a tautology. $a\ge b$ does not always imply $a=b$ , that is absurd. Also, the point of WLOG is that there are no cases, we only have to check one of the cases by symmetry. The reason I said $b\ge a$ rather than $b>a$ is to preserve the symmetry, so that the WLOG is valid.

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WolfusA

1900 posts

WolfusA

#10 h Jul 22, 2017, 7:09 AM • 2 Y

Y by Adventure10, Mango247

Sorry, but I made a mistake with this "tautology". I withdraw from accusations.

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Keith50

464 posts

Keith50

#11 h Jan 5, 2021, 6:27 AM

Y by

If $AC \neq BC$ , WLOG, assume that $\angle CAB > \angle CBA$ , then we have $BC > AC, BG > AE.$ Since $GF=DE$ , we have $FB=BG-GF >AE-GF=AE- DE=DA$ , so $GF=\frac{FB \cdot GC}{AB}=\frac{FB \cdot BC}{AB} > \frac{DA \cdot AC}{AB}=\frac{DA \cdot CE}{AB}=DE$ which contradicts $GF=DE$ , thus, $AC=BC$ . $\blacksquare$

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