Let $ABCD$ a convex quadrilateral with $AB=BC=CD$, with $AC$ not equal to $BD$ and $E$ be the intersection point of it's diagonals. Prove that $AE=DE$ if and only if $\angle BAD+\angle ADC = 120$.
Problem
Source: Balkan Mathematical Olympiad 2007, problem 1.
Tags: trigonometry, geometry, geometric transformation, reflection, trapezoid, circumcircle, trig identities
28.04.2007 14:46
"$\Rightarrow$" $AE = DE \Rightarrow \angle{BAD}+\angle{ADC}=120^\circ.$ : Denote $\angle{EAD}=\angle{EDA}=\beta$. From the sin law we have $\frac{AB}{\sin{\beta}}=\frac{AD}{\sin{ABD}}$ and $\frac{CD}{\sin{\beta}}=\frac{AD}{\sin{ACD}}$, but because $AC$ differs from $BD$ , we have that $\angle{ABD}=180-\angle{ACD}$, and denote $\angle{ACD}=x$. Thus, we have $\angle{BAC}=\angle{BCA}=x-2\beta$, therefore $\angle{CBD}=4\beta-x$, But $\angle{ACD}=x$, and $\angle{CED}=2\beta$, so $\angle{CDB}=180-x-2\beta$. But from $CD=CB$, we have that $\angle{CDB}=\angle{CBD}\iff 4\beta-x=180-x-2\beta \iff \beta=30$, and now just replacing it in $\angle{BAD}+\angle{ADC}=x-30+180-x-30=120$, we obtain the desired result. "$\Leftarrow$" $\angle{BAD}+\angle{ADC}=120^\circ \Rightarrow AE=DE$: Now let's denote $\angle{EAD}=\alpha$ and $\angle{EDA}=\beta$, again from the sin law we obtain that $\frac{\sin{\alpha}}{\sin{\beta}}=\frac{\sin{ACD}}{\sin{ABD}}$ $(\star)$. Denote now $\angle{CDB}=\angle{CBD}=x$, therefore $\angle{ACD}=180-\alpha-\beta-x$, so $\angle{ACB}=\alpha+\beta-x=\angle{CAB}\Rightarrow \angle{ABD}=180-2\alpha-2\beta+x$. But replacing the angles in $\angle{BAD}+\angle{ADC}=120$, we obtain $\alpha+\beta=60$, therefore replacing that in $(\star)$, we obtain that $\frac{\sin{\alpha}}{\sin{\beta}}=\frac{\sin{ACD}}{\sin{ABD}}=\frac{\sin{(60+x)}}{\sin{(120-x)}}=1$, therefore $\alpha=\beta \Rightarrow EA=ED$, and thus the problem is solved.
28.04.2007 19:40
$AE = DE \Rightarrow \angle{BAD}+\angle{ADC}=120^\circ. :$ $F$ is the reflection of $C$ through $AB$ We have $\angle FAD=\angle EAD=\angle EDA$ so $AF//BD$ But $DF=CD=AB$ and $AF$ is not equal to $BD$ so $ABDF$ is a isosceles trapezium So $\angle ABD+\angle ACD=\angle ABD+\angle AFD=180$ $\angle EAD+\angle EDA=\angle BEA=\angle DBC+\angle ACB$ so $\angle BAC+\angle CDA=2*(\angle DBC+\angle ACB)$ $(1)$ $\angle BAD+\angle CDA=360-\angle ABC-\angle BCD=180-\angle DBC-\angle ACB$ $(2)$ from (1) and (2) $\angle BAD+\angle CDA=120$ $\angle{BAD}+\angle{ADC}=120^\circ \Rightarrow AE=DE:$ $AB$ meets $CD$ at $M$ we have $\angle AMD=60$ $\angle AEC=180-\angle BEA$ $(3)$ $\angle AEC=\angle AMD+\angle MAC+\angle MDB=$$\angle60+\angle DBC+\angle ACB=60+\angle BEA$ $(4)$ from (3) and (4) $\angle BEA=60=\angle AMD$ so $B,M,C,E$ are cyclic Thus $\angle CDE=\angle CBE=\angle CME$ so triangle $MED$ is isosceles so $ED=EM$ similarly $EA=EM$ Thus $EA=ED$ $q.e.d$
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29.04.2007 00:26
Enunciation. Let $ABCD$ be a convex quadrilateral with $AB=BC=CD$ and let $E$ be the intersection point of it's diagonals. Prove that $\boxed{\ AE=DE\Longleftrightarrow BC\parallel AD\ \ \vee\ \ A+D= 120\ }$. Remark. Easy and very nice problem ! Proof. $\{\begin{array}{c}m(\widehat{DAC})=x\ ,\ m(\widehat{BDA})=y\\\\ AB=BC\Longleftrightarrow m(\widehat{CAB})=m(\widehat{BCA})=\alpha\\\\ BC=CD\Longleftrightarrow m(\widehat{DBC})=m(\widehat{CDB})=\beta\\\\ A+B+C+D=360^{\circ}\end{array}\|$ $\implies$ $\{\begin{array}{c}m(\widehat{ABD})=180^{\circ}-(x+y+\alpha )\\\\ m(\widehat{ACD})=180^{\circ}-(x+y+\beta )\\\\ x+y=\alpha+\beta\ ,\ \boxed{A+D=2(x+y)}\end{array}$. Apply the following well-known relation in the convex quadrilateral $ABCD$ ( memorize it ! ): $\boxed{\sin \widehat{ABD}\cdot\sin \widehat{DAC}\cdot\sin \widehat{CDB}\cdot\sin \widehat{BCA}=\sin \widehat{DBC}\cdot\sin \widehat{CAB}\cdot\sin \widehat{BDA}\cdot\sin \widehat{ACD}}$ $\Longleftrightarrow$ $\boxed{\sin (x+y+\alpha )\cdot\sin x=\sin y\cdot\sin (x+y+\beta )}$. Thus, $EA=ED$ $\Longleftrightarrow$ $x=y$ $\Longleftrightarrow$ $\{\begin{array}{c}\alpha+\beta =2x\\\\ \sin (\alpha+2x)=\sin (\beta+2x)\end{array}\|$ $\Longleftrightarrow$ $\{\begin{array}{c}\alpha+\beta =2x\\\ \alpha =\beta \ \ \vee\ \ \alpha+\beta+4x=\pi\end{array}$ $\Longleftrightarrow$ $\alpha =\beta =x\ \ \vee\ \ x=\frac{\pi}{6}$. Therefore, $EA=ED$ $\Longleftrightarrow$ $\{\begin{array}{cc}1\blacktriangleright & \alpha =\beta\Longleftrightarrow \boxed{\ BC\parallel AD\ }\\\\ 2\blacktriangleright & \alpha \ne\beta\Longleftrightarrow\{\begin{array}{c}x=y=\frac{\pi}{6}\ ,\ \alpha+\beta =\frac{\pi}{3}\\\\ A+D=120^{\circ}\end{array}\end{array}\|$ $\Longleftrightarrow$ $\boxed{\ BC\parallel AD\ \ \vee\ \ A+D=120^{\circ}\ }$. Remark. Denote $\boxed{\ \beta-\alpha =\delta\ }$. Thus, $\sin x\cdot\sin (x+y+\alpha )=\sin y\cdot\sin (x+y+\beta )$ $\Longleftrightarrow$ $\cos (y+\alpha )-\cos (2x+y+\alpha )=\cos (x+\beta )-\cos (x+2y+\beta )$ $\Longleftrightarrow$ $\cos (y+\alpha )-\cos (x+\beta )=\cos (2x+y+\alpha )-\cos (x+2y+\beta )$ $\Longleftrightarrow$ $\sin\frac{(x+y)+(\alpha+\beta )}{2}\cdot\sin \frac{x-y+\delta}{2}=\sin\frac{3(x+y)+(\alpha+\beta )}{2}\cdot\sin\frac{y-x+\delta }{2}$ $\Longleftrightarrow$ $\sin (x+y)\cdot\sin \frac{x-y+\delta }{2}=\sin 2(x+y)\cdot\sin \frac{y-x+\delta }{2}$ $\Longleftrightarrow$ $\boxed{\ \sin\frac{x-y+\delta }{2}=2\cos (x+y)\cdot\sin\frac{y-x+\delta }{2}\ }$. Therefore, $EA=ED$ $\Longleftrightarrow$ $x=y$ $\Longleftrightarrow$ $\sin\frac{\delta}{2}=2\cos 2x\cdot\sin\frac{\delta}{2}$ $\Longleftrightarrow$ $\delta =0\ (\alpha =\beta )\ \ \vee\ \ x=\frac{\pi}{6}$ a.s.o.
29.04.2007 00:29
Virgil Nicula wrote: Therefore, $EA=ED\iff\left\{\begin{array}{c}x=y=\frac{\pi}{6}\ ,\ \alpha+\beta =\frac{\pi}{3}\\\\ A+D=2(x+y)\end{array}\right\|\iff\boxed{A+D=120^{\circ}}$. Much better
29.04.2007 13:00
Who are the authors of BMO 2007 problems?
29.04.2007 19:23
I solved it during the competition completely synthetically...I don't have very much time now -i ll post my solution later.
29.04.2007 22:18
What if you consider the appropriate circles...
30.04.2007 10:49
Just one more approach, for $AE=DE \Rightarrow \angle A+\angle D = 120^{o}$: Triangles $AEB$ and $DEC$ are such that they have two equal sides and one eqaul angle, so $\angle ABD+\angle ACD = 180^{o}$ or $\angle ABD=\angle ACD$. In both cases, wanted equality follows almost directly. Bye
30.04.2007 14:41
I had the same idea that SpongeBob had for "$\Rightarrow$". Beautiful problem.
30.04.2007 17:32
SpongeBob wrote: Just one more approach, for $AE=DE \Rightarrow \angle A+\angle D = 120^{o}$: Triangles $AEB$ and $DEC$ are such that they have two equal sides and one eqaul angle, so $\angle ABD+\angle ACD = 180^{o}$ or $\angle ABD=\angle ACD$. In both cases, wanted equality follows almost directly. Bye Indeed, it is the same as my solution posted before, but with one observation, the case when $\angle{ABD}=\angle{ACD}$ doesn't work, because then $AC=BC$, so the only case available is when the angles are suplementary.
01.05.2007 17:11
I don't thnik so. When those angles are equal then this quad is cyclic, and there is nothing wrong with it.
01.05.2007 17:27
Yes, the quadrilateral is cyclic, but when $EA=ED$, then we obtain $AC=BD$, from the angle chasing, therefore this case doesn't work , because in the hypothesis we have that $AC$ differs from $BD$.
01.05.2007 19:40
IF two sides AB and DC extension and if they appointment in F, problem solved. Because $\angle CAB = \angle ACB= \alpha$ and $\angle DBC=\angle BDC = \beta$, and now we easy prove that $\alpha+\beta =60$ q.e.d. PS: if we suppose $\alpha+\beta$ is not equal with 60, we find #
02.05.2007 04:32
I think I have a different solution. I use undirected angles and segments. For convenience, denote $\angle BAE = \angle BCE = \alpha$ and $\angle CBE = \angle CDE = \beta$. We note that $\angle EAD+\angle EDA = \alpha+\beta$; hence it is sufficient to prove that the condition $\alpha+\beta = \pi/3$ is equivalent to the condition $AE = DE$. By the law of sines, \begin{eqnarray*}\frac{AE}{DE}&=& \frac{AE}{EB}\cdot \frac{EB}{EC}\cdot \frac{EC}{DE}\\ &=& \frac{\sin ABE}{\sin EAB}\cdot \frac{\sin ECB}{\sin EBC}\cdot \frac{\sin EDC}{\sin DCE}\\ &=& \frac{\sin (\pi-2\alpha-\beta)}{\sin \alpha}\cdot \frac{\sin \alpha}{\sin \beta}\cdot \frac{\sin \beta}{\sin (\pi-2\beta-\alpha)}\\ &=& \frac{\sin (\pi-2\alpha-\beta)}{\sin (\pi-2\beta-\alpha)}. \end{eqnarray*} Thus $AE = DE$ if and only if $\sin (\pi-2\alpha-\beta) = \sin (\pi-2\beta-\alpha)$. Since we know $\alpha \neq \beta$ (or else triangles $ABC, BCD$ would be congruent and we would have $AC= BD$) and $0 < \alpha, \beta < \pi/2$, this later condition is equivalent to the condition $\pi-2\alpha-\beta = \pi-(\pi-2\beta-\alpha)$, or $3(\alpha+\beta) = \pi$, which gives us what we wanted.
02.05.2007 10:25
pohoatza wrote: Yes, the quadrilateral is cyclic, but when $EA=ED$, then we obtain $AC=BD$, from the angle chasing, therefore this case doesn't work , because in the hypothesis we have that $AC$ differs from $BD$. Oh... Sorry, I didn't see that in statement of a problem, and I really don't know why did they put it there, maybe to avoid giving 2 or 3 points on trivial cases. Bye
07.05.2007 22:57
If $AB$ and $CD$ meet at P: - BPDE is cyclic - E is the circumcentre of APD - the circumcircles of ABE,BPC,CED have the same radius - the circumcircles of ABE,CED meet on AD These are some of the useless things I proved during the competition By the way, nice problem.
08.05.2007 14:33
edriv these are not useless...don't u remember?my solution uses the equality of one pair of circles u proved..
26.08.2010 20:20
I think that this solution is different. Solution First note that the condition $AE=DE$ is equivalent to $\angle{DAC}=\angle{ADB}$. Let $P$ be a point external to $ABCD$ such that $ACDP$ is a parallelogram. Since $AP=CD=AB$, $\angle{PDA}=\angle{DAC}=\angle{ADB}$ is equivalent to $DA$ bisects $\angle{PDB}$ and meets the perpendicular bisector of $BP$ at $A$. This occurs if and only if $ABDP$ is cyclic. Let $x=\angle BAD+\angle ADC = 120$. Now note that since triangles $ABC$ and $BCD$ are isosceles, \[\angle{PDB}=\angle{DAC}+\angle{ADB}=\angle{DBC}+\angle{ACB}=\frac{x}{2}\] Now, $ABDP$ is cyclic if and only if $180=\angle{PDB}+\angle{PAB}=\frac{3x}{2}$ or $x=120$.
23.11.2014 19:17
If: extend $AC$ to $F$ such that $AD=FD$. Angle chase to get the relationship between triangles $ABD$ and $DCF$ is SSA, so either $\angle BDA= \angle BDC$ or $\angle CAD= 30$. The former gives $AC=BD$, which is a contradiction. The latter gives the required. Only if: now, extend $AC$ to $F$ such that $\angle BAD= \angle CDF$. Again, angle chase to get that $AD=FD$ and $\angle FAD+\angle AFD=60$ which gives the required.
04.02.2016 12:16
Note that $AFB,DFC$ have same radii and hence if $AE=AD$ then either $\angle{ABF}=\angle{CFD}$ or their sum is $180$.One gives contradiction to $BD\neq AC$ while other solves the problem
28.09.2016 18:05
Here is my solution Let's say that $\measuredangle BAC=\measuredangle BCA=\beta$ and $\measuredangle DBC=\measuredangle BDC=\alpha$ Let's first assume $AE=DE$ So $\measuredangle DAE=\measuredangle ADE=\frac {\alpha+\beta}{2}$ Let's write The law of Sines in triangle $ABE$ $\frac{AE}{\sin(2\beta+\alpha)}=\frac{AB}{sin(\alpha+\beta)}$ $(1)$ Now We ll write The law of Sines in triangle $DCE$ $\frac{DE}{\sin(2\alpha+\beta)}=\frac{CD}{\sin(\alpha +\beta)}$ $(2)$ Combining ($1$) and ($2$) We get $\sin(2\alpha+\beta)=\sin(2\beta+\alpha)$ from here two cases exist. $i)$ $\alpha=\beta$ which means Diagonals are equal contradiction. $ii)$ $\alpha+\beta=60^{\circ}$ Done!!!
28.09.2016 18:11
Let's second assume $\measuredangle BAC+\measuredangle ADB=120^{\circ}$ this gives us $\alpha+\beta=60^{\circ}$ İf we write The law of Sines in triangles $ABE$ and $DCE$ we ll need to prove that $\sin(2\alpha+\beta)=\sin(2\beta+\alpha)$ $i)$ $\alpha=\beta$ which means Diagonals are equal contradiction. $ii)$ $3(\alpha+\beta)=180^{\circ}$ or $\alpha+\beta=60^{\circ}$ DONE!!!
20.05.2020 13:55
Let $\angle ACB=x, \angle DBC=y$. From isosceles triangle $ACB$ and sine law: $\frac{AC}{\sin 2x}=\frac{BC}{\sin x} \implies AC=2BC \cdot \cos x$ From triangle $CEB$ and sine law: $\frac{CE}{\sin y}=\frac{CB}{\sin(x+y)} \implies CE=\frac{BC \cdot \sin y}{\sin(x+y)}$ Now: $AE=\frac{\sin(2x+y)}{\sin(x+y)}$ and $DE=\frac{\sin(2y+x)}{\sin(x+y)} \implies$ $\frac{\sin(2x+y)}{\sin(x+y)}=\frac{\sin(2y+x)}{\sin(x+y)}$ or $\sin(2x+y)-\sin(2y+x)=0$ and $2\sin(\frac{x-y}{2}) \cos(\frac{3x+3y}{2})=0$ Now $3x+3y=180^\circ$ and $x+y=60^\circ$ $\angle ABC+\angle BCD=240^\circ$ $\implies \angle BAD+\angle ADC=120^\circ$
03.02.2024 11:49
Ceva sine at quadrilateral
12.04.2024 16:50
$i)$ $AE = DE \Longrightarrow \angle BAD + \angle ADC = 120^\circ$ Let $AB=BC=CD=a$, $BE=x$, $CE=y$, $AE=ED=z$. From Stewart's theorem in $\triangle ABC$ and $\triangle BCD$, we have $a^2=yz+x^2=xz+y^2$. Rearranging gives $x^2-y^2=z(x-y)$. Since $x\neq y$ by the condition of the problem, we obtain $x+y=z$. Let $M$ be the reflection of $E$ across the midpoint of $AC$, and let $N$ be the reflection of $D$ across the midpoint of $AD$. Then, $AM=CE=y$, $BM=BE=x$, $DN=BE=x$, and $CN=CE=y$. Also, $ME=x$ and $NE=y$. Therefore, $\triangle BEM$ and $\triangle CEN$ are equilateral triangles. From here, the result can be obtained with simple angle calculations. Using $\triangle CND \cong \triangle AMB$, we have $\angle BAD + \angle CDA = \angle BAM + \angle EAD + \angle EDA+ \angle NDC = \angle BAM + \angle BEA + \angle MBA = \angle MBE + \angle BEA = 60^\circ + 60^\circ=120^\circ$. $ii)$ $\angle BAD + \angle ADC = 120^\circ \Longrightarrow AE=DE$ Let $F$ be the intersection of $AB$ and $DC$, with $\angle AFD=60^\circ$. If we let $\angle BAC= \angle BCA=\alpha$, then $\angle FBC=2\alpha$, $\angle FCB=120^\circ-2\alpha$, and $\angle CBD = \angle CDB=60^\circ-\alpha$. Taking $M$ on $[AE]$ such that $BE=EM$, we have $\triangle BEM$ as an equilateral triangle, and by similarity or symmetry, $AM=CE$. Thus, $AE=AM+ME=CE+BE$. Similarly, taking $N$ on $[DE]$ gives us $DE=CE+BE=AE$, we obtain the desired result.