Prove that $\angle FAC = \angle EDB$

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Prove that $\angle FAC = \angle EDB$

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Source: All-Russian Olympiad 1996, Grade 10, First Day, Problem 1

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172 posts

#1 h Apr 18, 2013, 7:06 PM • 1 Y

Y by Adventure10

Points $E$ and $F$ are given on side $BC$ of convex quadrilateral $ABCD$ (with $E$ closer than $F$ to $B$ ). It is known that $\angle BAE = \angle CDF$ and $\angle EAF = \angle FDE$ . Prove that $\angle FAC = \angle EDB$ .

M. Smurov

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1479 posts

#2 h Apr 18, 2013, 7:39 PM • 5 Y

Y by vsathiam, Understandingmathematics, Adventure10, SomeonecoolLovesMaths, and 1 other user

$\angle{ADC}+\angle{ABC}=\angle{FDA}+\angle{CDF}+\angle{AEF}-\angle{BAE}=\pi$ , done.

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#3 h Nov 22, 2014, 11:50 PM • 3 Y

Y by arulxz, arinastronomy, Adventure10

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#4 h Nov 30, 2016, 5:11 PM • 1 Y

Y by Adventure10

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#5 h May 4, 2019, 11:13 AM • 2 Y

Y by Adventure10, Mango247

Solution

This post has been edited 2 times. Last edited by thegreatp.d, May 4, 2019, 11:14 AM
Reason: Typo

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#10 h Jul 23, 2020, 11:15 AM • 1 Y

Y by MrCriminal

micliva wrote:

Points $E$ and $F$ are given on side $BC$ of convex quadrilateral $ABCD$ (with $E$ closer than $F$ to $B$ ). It is known that $\angle BAE = \angle CDF$ and $\angle EAF = \angle FDE$ . Prove that $\angle FAC = \angle EDB$ .

M. Smurov

$\angle EAF = \angle EDF$
So, quadrilateral AEFD is cyclic
$\angle ADE = \angle AFE$
$\angle ADC = \angle ADE$ + $\angle EDF + \angle FDC$ ..... (1)
$\angle ABC = 180- \angle AFB$ - $\angle BAE - \angle EAF$ ....(2)
(1)+(2),
$\angle ABC + \angle ADC$ = 180
So, ABCD is cyclic.
So, $\angle BAC = \angle BDC$
Or, $\angle FAC + \angle BAF$ =
$\angle BDF + \angle EDB$
Or, $\angle FAC = \angle EDB$

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270 posts

#11 h Oct 12, 2020, 6:41 AM • 1 Y

Y by ASLMATH

just continue $AF$ and $DE$ to intersect $DC$ and $DB$ at $Q$ and $P$ ,

then $ADEF$ and $ADQP$ are cyclic.
then easily prove that $ACQ \sim DBP$
DONE :roll:

:roll:

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1574 posts

#12 h Nov 16, 2020, 11:11 PM

Y by

Note that quadrilateral $AEFD$ is cyclic, which implies $\angle FDA=180^\circ-\angle FEA=\angle BEA$ . This also implies that $\angle EBA=180^\circ-\angle BAE-\angle BEA$ . However, we note that $\angle CDA=\angle CDF+\angle FDA=\angle BAE+\angle BEA$ which implies $ABCD$ is cyclic. To finish, we note that $\angle BAC=\angle CDB$ which implies $\angle FAC=\angle EDB$ . $\blacksquare$

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#13 h Sep 30, 2021, 4:22 AM

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how is russia so good.

Solution

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#14 h Sep 30, 2021, 4:40 AM • 1 Y

Y by Truly_for_maths

OlympusHero wrote:

how is russia so good.

Solution

Funny how one dollar sign can make everything so much better.

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#15 h Dec 19, 2021, 11:02 AM

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∠EAF = ∠FDE ---> EADF is cyclic.
∠FCD = 180 - ∠CDF - ∠DFC = 180 - ∠EAB - ∠DAE = 180 - DAB ---> ABCD is cyclic.
∠CDB = ∠CAB ---> ∠EDB = ∠CAF

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ajax31

#16 h Dec 28, 2021, 12:50 AM

Y by

$AEFD$ is cyclic because $\angle EAF=\angle FDE$ .
Let $\angle DAF=z, \angle BAE=x, \angle EAF=y$ . We then proceed to angle chase. $\angle EFD=180-y-z$ because $AEFD$ is cyclic and opposite angles in a cyclic quadrilateral add to $180$ .
Next, $\angle CFD=y+z$ by supplementary angles and $\angle FCD=180-(x+y+z)$ because the angles in a triangle add up to $180$ .
Therefore, ABCD is cyclic because $\angle A+\angle C=180$ , which shows that $\angle BDC=\angle BAC$ . Finally, $\angle FAC=\angle EDB$ because you just subtract equal angles $\angle BAF$ and $\angle EDC$ . $\square$

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#17 h Jul 28, 2022, 1:22 AM • 1 Y

Y by Mango247

Since $\angle{BAE}=\angle{CDF}$ and $\angle{EAF}=\angle{FDE}$ , if we add those equalities to the desired conclusion, we then need to prove that $ABCD$ is cyclic.

From $\angle{EAF}=\angle{FDE}$ , we know that $AEFD$ is cyclic, so let $\angle{DFE}=x$ and $\angle{DAE}=180-x$ . Let $\angle{BAE}=\angle{CDF}=y$ . We then have $\angle{DAB}=180-x+y$ , and by sums of angles in triangle we have $\angle{DCB}=180-(180-x+y)$ , so $ABCD$ is cyclic as desired. $\blacksquare$

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#18 h Oct 10, 2022, 5:34 PM

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Since $\angle{EAF}=\angle{FDE},~ AEFD~ \text{is cyclic}. ~\angle{EAD}=\angle{DFC}.~\angle{BAD}=\angle{BAE}+\angle{EAD}=\angle{CFD}+\angle{DFC}=180^\circ-\angle{DCB} \Rightarrow ~ABCD ~\text{is cyclic.} \Rightarrow \angle{BAC}=\angle{BCD}. ~\text{Since } \angle{BAE}=\angle{CDF} ~\text{and}~ \angle{EAF}=\angle{EDF} \Rightarrow \angle{FAC}=\angle{EDB}$ as required. This was a lot easier in hindsight.

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#19 h Dec 19, 2022, 8:48 PM • 3 Y

Y by Mango247, Mango247, Mango247

how did that work // half hour

Observe that since $\angle EAF=\angle FDE$ , $EADF$ is cyclic, wherefore
$\begin{align*} \angle BAD+\angle BCD&=\angle BAE+\angle DAE+\angle FCD\\ &=\angle CDF+\angle DAE+\angle FCD\\ &\stackrel{\triangle CDF}{=}\angle DFE+\angle DAE\\ &\stackrel{(EADF)}{=}180^\circ. \end{align*}$ So $ABCD$ is cyclic as well, yielding $\angle BAC=\angle CDE$ or
$\angle FAC=\angle BAC-\angle BAE-\angle EAF=\angle CDB-\angle CDF-\angle FDE=\angle EDB,$ as desired. $\blacksquare$

https://www.geogebra.org/calculator/b2cxmhfh

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#20 h Apr 21, 2023, 5:29 PM

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This is equivalent to showing that $ABCD$ is cyclic as the condition is equivalent to $\measuredangle CDB = \measuredangle CAB$ . Quadrilateral $ADEF$ is cyclic since $\measuredangle EAF = \angle EDF$ .
Note $\measuredangle ABC = \measuredangle ABE = \measuredangle BAE + \measuredangle AEB = \measuredangle BAE + \angle AEF$ and that $\measuredangle ADC = \angle ADF + \angle FDC$ Finally, $\measuredangle ABC = \measuredangle BAE + \measuredangle AEF = \measuredangle BAE + \measuredangle ADF = \measuredangle FDC + \measuredangle ADF = \measuredangle ADC$ $[asy] pair a, b, c, d, e, f; a = dir(140); d = dir(220); f = dir(300); e = dir(30); b = 1.3 * e - 0.3 * f; c = intersectionpoints((100*e-99*f)--(100*f-99*e),circumcircle(a,d,b))[1]; draw(a--e--f--d--cycle,orange); draw(d--c--f,blue); draw(e--b--a,blue); draw(circumcircle(a, b, c),blue); draw(circle((0,0),1),orange); dot("$A$", a, dir(120)); dot("$D$", d, dir(240)); dot("$F$", f, dir(330)); dot("$E$", e, dir(30)); dot("$B$", b, dir(30)); dot("$C$", c, dir(270)); [/asy]$

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#21 h Jun 16, 2023, 8:32 AM

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Since $\angle EAF = \angle FDE, FDAE$ is cyclic. Because $\angle DAF = \angle DEF, \angle EAF = \angle FDE, \angle BAE, \angle CDF,$ $\angle DCE = 180^\circ - \angle EAF - \angle BAE.$ $\angle DCE + \angle BAD = 180^\circ$ implies that $ABCD$ is cyclic. Hence, $\angle BDC = \angle CAB \implies \angle FAC = \angle EDB.$ $\blacksquare.$

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#22 h Jul 17, 2023, 10:35 PM

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Solution: Clearly, right off the bat, we have that $EFDA$ is cyclic. Because of this, we can set $\angle{DEF}=\angle{DAF}=\alpha$ and $\angle{AFE}=\angle{ADE}=\beta$ . Now, setting $\angle{AED}=\angle{AFD} = \theta$ and $\angle{BAE}=\angle{FDC}=\phi$ , we can angle chase to get that $\angle{B}=\theta + \alpha - \phi$ and $\angle{D} = 180 - \alpha - \theta + \phi$ . Clearly, $\angle{B} + \angle{D} = 180^\circ$ , so $ABCD$ is cyclic. Finally, we have $\angle{CAB}=\angle{BDC}$ , which implies that $\angle FAC = \angle EDB$ . We are done. $\blacksquare$

Comments: From EGMO chapter 1. Pretty simple problem.

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#23 h Jan 19, 2024, 4:06 PM

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Let $G$ be the intersection of $AF$ and $DE.$ Since $\angle EAF=\angle EDF,$ $AEFD$ is cyclic. As such, $\angle ADE=\angle AFE,$ and $\angle DAF=\angle DEF.$ We also have $\angle DGF=180^{\circ} - \angle AGD=180^{\circ} -\left(180-\left(\angle DAF+\angle ADE\right)\right)=\angle DAF+\angle ADE,$ so $\angle GFD=180^{\circ}-\angle DGF-\angle EDF=180^{\circ}-\angle DAF-\angle ADE-\angle EAF.$ As such, $\angle DFC=180^{\circ}-\angle DFE=180^{\circ}-\left(\angle GFD+\angle AFE\right)=180^{\circ}-\left(180^{\circ}-\angle DAF-\angle ADE-\angle EAF+\angle AFE\right)=180^{\circ}-\left(180^{\circ}-\angle DAF-\angle EAF\right)=\angle DAF+\angle EAF.$ As such, $\angle DCB=\angle DCF=180^{\circ}-\left(\angle CDF+\angle DFC\right)=180^{\circ}-\left(\angle BAE+\angle DAF+\angle EAF\right)=180^{\circ}-\angle BAD,$ so $\angle DCB+\angle BAD=180^{\circ},$ and $ABCD$ is cyclic. As such, $\angle BAC=\angle BDC,$ so $\angle FAC=\angle BAC-\angle BAE-\angle EAF=\angle BDC-\angle CDF-\angle FDE=\angle EDB,$ as desired. $\Box$

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#24 h Jan 25, 2024, 2:11 AM

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Since $\angle EAF = \angle FDE$ , we know that $ADFE$ is cyclic.

SInce $\angle BAF = \angle BAE + \angle EAF = \angle CDF + \angle FDE = \angle CDE$ , we just need to prove that $\angle BAC = \angle BDC$ or that $ABCD$ is cyclic

Since $\angle ABC = \angle AEC - \angle EAB = 180^\circ - \angle ADF - \angle CDF = 180^\circ-\angle ADC$ , $ABCD$ is cyclic, as desired.

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blueberryfaygo_55

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#25 h Mar 16, 2024, 2:40 PM

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Since $\angle FDE = \angle FAE$ , $ADFE$ is a cyclic quadrilateral. It follows that $\begin{align*} \angle DAE &= 180^{\circ} - \angle DFE \\ &= \angle DFC \\ &= 180^{\circ} - \angle CDF - \angle DCF \end{align*}$ However, we also have $\angle DAE = \angle DAB - \angle BAE$ . Thus, $180^{\circ} - \angle CDF - \angle DCF = \angle DAE = \angle DAB - \angle BAE$ and $\angle EAB = \angle CDF$ , so it follows that $180^{\circ} - \angle DCF = \angle DAB$ or $ABCD$ is cyclic. Then, $\angle CDB = \angle CAB$ , but we have $\angle CDE = \angle BAF$ , so $\angle CAF = \angle BDE$ as desired.

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Aaronjudgeisgoat

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Aaronjudgeisgoat

#26 h Jun 7, 2024, 1:18 AM

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Note that $\angle FAC = \angle EDB$ if and only if $ABCD$ is cyclic.

Let us say that $\angle BAE = \angle CDF = x, \angle EAF = \angle EDF =y,$ and $\angle EDA=z.$

Since $\angle EAF=\angle EDF, EFDA$ is cyclic. This means that $\angle EFA = \angle EDA = z.$ Therefore, $\angle ABC= 180-x-y-z,$ which means that $\angle ABC +\angle CDA=180,$ proving that $ABCD$ is cyclic, as desired.

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G0d_0f_D34th_h3r3

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G0d_0f_D34th_h3r3

#27 h Jun 9, 2024, 8:23 AM

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In quadrilateral $AEFD, \angle EAF = \angle EDF$ . So, quadrilateral $AEFD$ is cyclic.
So, $\angle DFC = \angle EAF + \angle FAD$ .
In $\Delta DFC, \angle FCD = 180^{\circ} - \angle DFC - \angle CDF \\ = 180^{\circ} - (\angle EAF + \angle FAD) - \angle CDF \\ = 180^{\circ} - (\angle EAF + \angle FAD) - \angle BAE$ .
Then quadrilateral $ABCD$ is cyclic. Hence $\angle BAC = \angle BDC \\ \Rightarrow \angle BAC - \angle BAF = \angle BDC - \angle CDE \\ \Rightarrow \angle FAC = \angle EDB$ .

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#28 h Aug 26, 2024, 3:30 PM

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$AEFD$ is a cyclic quadrilateral. Take point $K$ on line $AD$ such that $KA>KD$ . Then by angle chasing,
$\angle ABC=\angle AEC-\angle EAB=\angle FDK-\angle FDC=\angle CDK$ which implies $ABCD$ quadrilateral being cyclic. Thus
$\angle FAC=\angle FAD-\angle CAD=\angle FED-\angle FBD=\angle EDB$ as desired.

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#29 h Jan 10, 2025, 8:23 AM

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Let $\angle BAE = \angle FDC = x$ . Since $AEFD$ is cyclic, $\angle DFE + \angle DAE = \angle AEF + \angle ADF = 180$ . $\angle DFE + \angle DFC$ and $\angle AEF + \angle AEB$ also equal to $180$ so $\angle DAE = \angle DFC$ and $\angle ADF = \angle AEB$ . $x+\angle AEB+\angle ABE = x+\angle ADF + \angle ABC = \angle ADC +\angle ABC = 180$ so $ABCD$ is cyclic. Hence $\angle BAC = \angle BDC$ so $\angle FAC = \angle BDE$

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#30 h Feb 17, 2025, 1:32 PM

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Angle chase to obtain $ABCD$ cyclic, then angle chase some more.

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#31 h Yesterday at 5:14 AM

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Since $\angle{EAF} = \angle{EDF}$ , we know that $AEFD$ is cyclic. This also means that $\angle{FDA} = 180^{\circ} - \angle{FEA}$ , implying $\angle{FDA} = \angle{BEA}$ .

We also have $\angle{ABE} = 180^{\circ} - \angle{BEA} - \angle{BAE}$ . If we substitute in angles ( $\angle{BEA} = \angle{FDA}$ from earlier and $\angle{BAE} = \angle{FDC}$ from the given), we find that $\angle{ABE} = 180^{\circ} - \angle{FDA} - \angle{FDC} = 180^{\circ} - \angle{CDA}$ . Thus, $ABCD$ is cyclic, as the opposite angles in the quadrilateral are supplementary.

We then have that $\angle{BAC} = \angle{BDC}$ by the cyclic condition, and subtracting off some equal angles yields $\angle{FAC} = \angle{BDE}$ , as desired.

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