Parallelogram problem

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Source: INMO 1997 Problem 1

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Rushil

#1 h Oct 6, 2005, 2:33 PM • 1 Y

Y by Adventure10

Let $ABCD$ be a parallelogram. Suppose a line passing through $C$ and lying outside the parallelogram meets $AB$ and $AD$ produced at $E$ and $F$ respectively. Show that $AC^2 + CE \cdot CF = AB \cdot AE + AD \cdot AF .$

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yetti

#2 h Oct 10, 2005, 5:47 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Denote $\alpha = \angle DAB,\ \beta = \angle ABC$ the parallelogram internal angles, $\alpha + \beta = 180^\circ$ . Using the cosine theorem, for example, for the triangle $\triangle FDC$ together with $AB = DC$ , we get

$CF = DC^2 + DF^2 - 2\ DC \cdot DF \cos \alpha = AB^2 + DF^2 - 2\ AB \cdot DF\ \cos \alpha$

$1= \left(\frac{AB}{CF}\right)^2 + \left(\frac{DF}{CF}\right)^2 - 2\ \frac{AB}{CF} \cdot \frac{DF}{CF}\ \cos \alpha$

The triangles $\triangle FDC \sim \triangle CBE$ , because their corresponding side lines are parallel: $DC \parallel AB \equiv BE,\ DF \equiv AD \parallel BC$ and $CF \equiv CF$ . Hence,

$\frac{AB}{CF} = \frac{DC}{CF} = \frac{BE}{CE}$

$\frac{DF}{CF} = \frac{BC}{CE} = \frac{AD}{CE}$

Substituting appropriately (i.e., not everywhere) for $\frac{AB}{CF},\ \frac{DF}{CF}$ into the last cosine theorem formula,

$1 = \frac{AB}{CF} \cdot \frac{BE}{CE} + \frac{AD}{CE} \cdot \frac{DF}{CF} - 2\ \frac{AB}{CF} \cdot \frac{AD}{CE}\ \cos \alpha$

Multiplying by the product $CE \cdot CF$ and substituting for $BE = AE - AB,\ DF = AF - AD$ ,

$CE \cdot CF = AB \cdot (AE - AB) + AD \cdot (AF - AD) - 2\ AB \cdot AD\ \cos \alpha$

$CE \cdot CF = AB \cdot AE + AD \cdot AF - (AB^2 + AD^2 + 2\ AB \cdot AD\ \cos \alpha)$

Finally, using the cosine theorem for the triangle $\triangle ABC$ together with $AD = BC,\ \cos \alpha = -\cos \beta$ , we get

$AC^2 = AB^2 + BC^2 - 2\ AB \cdot BC\ \cos \beta = AB^2 + AD^2 + 2\ AB \cdot AD\ \cos \alpha$

Substituting this to the previous equation yields the desired result:

$CE \cdot CF = AB \cdot AE + AD \cdot AF - AC^2$

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#3 h Oct 10, 2005, 2:05 PM • 2 Y

Y by Adventure10, Mango247

Rushil wrote:

Let $ABCD$ be a parallelogram. Suppose a line passing through $C$ and lying outside the parallelogram meets $AB$ and $AD$ produced at $E$ and $F$ respectively. Show that $AC^2 + CE \cdot CF = AB \cdot AE + AD \cdot AF .$

Let's try to find a solution hiding the fact that this problem is nothing but a reformulation of Stewart's theorem...

Let the line AC meet the circumcircle of triangle AEF at G (apart from A). Then, by the intersecting chords theorem, $CE\cdot CF=AC\cdot CG$ .
Let the line AC meet the circumcircle of triangle EBC at U (apart from C). Then, by the intersecting chords theorem, $AB\cdot AE=AU\cdot AC$ .
Let the line AC meet the circumcircle of triangle FCD at V (apart from C). Then, by the intersecting chords theorem, $AD\cdot AF=AV\cdot AC$ .

Hence, the equation that we have to prove, $AC^2 + CE \cdot CF = AB \cdot AE + AD \cdot AF$ , becomes $AC^2+AC\cdot CG=AU\cdot AC+AV\cdot AC$ . Division by AC transforms this into AC + CG = AU + AV.

Now, we will show that AU = VG. This will yield AC + CG = AG = VG + AV = AU + AV, and thus the problem will be solved.

The rest of the solution is angle chasing, so we will use directed angles modulo 180°.

Since the points E, B, C and U lie on one circle, < CUB = < CEB and < UBE = < UCE. In other words, < AUB = - < AEC and < UBA = - < ECA. Hence, the triangles AUB and AEC are oppositely similar, so that $\frac{AU}{AE}=\frac{AB}{AC}$ .
Since the quadrilateral ABCD is a parallelogram, BC || AD, so that < EAF = < EBC. Since the points A, E, F and G lie on one circle, < EFG = < EAG and < EGF = < EAF. In other words, < EFG = - < CAB and < EGF = - < CBA (in fact, since < EAF = < EBC, the equation < EGF = < EAF becomes < EGF = < EBC = - < CBA). Hence, the triangles FGE and ABC are oppositely similar, so that $\frac{AB}{AC}=\frac{FG}{FE}$ .
Since the points F, C, D and V lie on one circle, < CVF = < CDF. Since the quadrilateral ABCD is a parallelogram, CD || AB, and thus < CDF = < EAF. Hence, < GVF = < CVF = < CDF = < EAF. Also, since the points A, E, F and G lie on one circle, < AGF = < AEF, what is equivalent to < VGF = < AEF. From < VGF = < AEF and < GVF = < EAF, it follows that the triangles VGF and AEF are directly similar, so that $\frac{FG}{FE}=\frac{VG}{AE}$ .

Hence, $\frac{AU}{AE}=\frac{AB}{AC}=\frac{FG}{FE}=\frac{VG}{AE}$ , so that AU = VG, and we are done.

darij

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#4 h Feb 18, 2006, 9:12 AM • 4 Y

Y by satej, Vasu090, Adventure10, Mango247

$BC\parallel AF\Longrightarrow \frac{AE}{AB}=\frac{FE}{FC}\Longrightarrow AB=\frac{AE\cdot FC}{EF}\ \ (1)\ .$

$CD\parallel AE\Longrightarrow \frac{AF}{AD}=\frac{EF}{EC}\Longrightarrow AD=\frac{AF\cdot EC}{EF}\ \ (2)\ .$

The Stewart's theorem in the triangle $AEF$ for the cevian-line $AC\Longrightarrow$

$AC^2\cdot EF+EF\cdot EC\cdot FC=AE^2\cdot CF+AF^2\cdot CE\Longrightarrow$

$AC^2+CE\cdot CF=AE\cdot \frac{AE\cdot FC}{EF}+AF\cdot \frac{AF\cdot EC}{EF}\ .$

From the relations $(1)$ and $(2)$ results: $AC^2+CE\cdot CF=AB\cdot AE+AD\cdot AF\ .$

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#5 h Aug 6, 2023, 4:27 AM

Y by

By Thales theorem on ∆ EAF, BC|| AD
EB/ AB=CE/ CF
AE/ AB=EF/ CF( ADD 1 O BOTH SIDES)....(1)
By Thales theorem on ∆ EAF, DC||AE
ED/ AD=CF/ CE
AF/ AD=EF/ CE( ADD 1 O BOTH SIDES)....(2)
By Stewart theorem in ∆ EAF,
CE.CF.EF+AC².EF=AF².CE+AE².CF
AC²+CE.CF=AF²(CE/EF)+ AE²(CF/EF)
AC²+CE.CF=AF².(AD/ AF)+ AE²(AB/AE)[ From (1)&(2)]
AC²+CE.CF=AD.AF+ AB.AE( proved)
# Krishijivi

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