In the convex quadrilateral $ABCD$ we have $\angle B = \angle C$ and $\angle D = 90^{\circ}.$ Suppose that $|AB| = 2|CD|.$ Prove that the angle bisector of $\angle ACB$ is perpendicular to $CD.$
Problem
Source: Benelux Mathematical Olympiad 2017, Problem 3
Tags: geometry, Benelux
Blast_S1
07.05.2017 02:27
First, extend $\overline{AB}$ and $\overline{DC}$ to meet at point $E$, and draw $\overline{AC}$
WLOG, let $CD=x, AB=2x, m\angle ACD=\phi, m\angle ABC=m\angle DCB=\theta$
$CLAIM:$ $EC=EB$
$PROOF:$
This is simple angle chasing, $m\angle ECB=m\angle EBC=180^\circ-\theta$
so $\triangle EBC$ is isosceles, and therefore $EB=EC$
we can now do a bit more angle chasing to get that $m\angle DAC=90^\circ-\phi$ and $m\angle ACB=\theta-\phi$
Consider $\triangle AEC$, we note that $\frac{AC}{EC}=\frac{\sin\angle AEC}{\sin\angle EAC}$ by a simple LOS
applying ratio lemma yields
$$\frac{AB}{BE}=\frac{2x}{BE}=\frac{AC\cdot\sin\angle ACB}{CE\cdot\sin\angle ECB}=\frac{\sin\angle AEC\cdot\sin\angle ACB}{\sin\angle EAC\cdot\sin\angle ECB}$$
similarly, for $\triangle AED$, we have $\frac{AD}{AE}=\frac{\sin\angle AEC}{\sin\angle ADC}$
however this time, since $m\angle ADE=90^\circ$, the above fraction simply reduces to $\sin\angle AEC$
now, using ratio lemma again, we have
$$\frac{DC}{CE}=\frac{x}{CE}=\frac{AD\sin\angle DAC}{AE\sin\angle EAC}=\frac{\sin\angle AEC\sin\angle DAC}{\sin\angle EAC}$$
finally, since $BE=CE$, therefore $\frac{2x}{BE}=2\cdot\frac{x}{CE}$, or
$$\frac{\sin\angle AEC\cdot\sin\angle ACB}{\sin\angle EAC\cdot\sin\angle ECB}=2\cdot\frac{\sin\angle AEC\sin\angle DAC}{\sin\angle EAC}\iff \sin\angle ACB=2\cdot\sin\angle ECB\cdot\sin\angle DAC$$or
$$\sin(\theta-\phi)=\sin(180^\circ-\theta)\sin(90^\circ-\phi)$$$$\sin\theta\cos\phi-\sin\phi\cos\theta=2\sin\theta\cos\phi$$$$0=\sin\theta\cos\phi+\sin\phi\cos\theta$$$$0=\sin(\theta+\phi)$$the only possible way for this to work is if $\theta+\phi=180^\circ\iff \phi=180^\circ-\theta$
this yields that $m\angle ACD=m\angle BCE=180^\circ-\theta$, so the angle bisector of $\angle BCA$ is clearly also the angle bisector of $\angle ECD$, which is a straight angle
and we're done bashng
PROF65
07.05.2017 03:03
consider $E$ the symmetric of $C$ in $D $, $KEA,KCB$ are isoceles where $K\equiv DC\cap AB$ .$ \angle DCA= \angle CDA =\pi -\angle C \implies \frac{1}{2}\angle ACB=\angle C-\frac{\pi}{2} $ which means that the bisector of $\angle ACB$ is perpendicular to$DC$. RH HAS
CFOP
07.05.2017 03:45
Let the angle bisector of $\angle ACB$ intersect $\overline{AB}$ at $E$.
Construct point $M$ on $\overline{AB}$ such that $\overline{AD}\parallel\overline{BC}$.
This makes $MBCD$ an isosceles trapezoid with $MB=CD$. Thus, $AM=MB$.
Since $\overline{AD}\parallel\overline{BC}$, it follows that $AP=PC$ where $P$ is the intersection of $\overline{MD}$ and $\overline{AC}$.
In right $\triangle ACD$, $\overline{DP}$ is a median which means that $CP=DP$.
Let $m\angle BCE=m\angle ECP=x$ and $m\angle CDP=y$.
Since $MBCD$ an isosceles trapezoid, $m\angle BCD=180-y$, $m\angle ECP=180-x-y$, and $m\angle ACD=180-2x-y$.
Using the fact that $CP=DP$, $m\angle ACD=m\angle CDP \implies 180-2x-y=y \implies 2x+2y=180 \implies x+y=90$.
Plugging this into back $m\angle ECP=180-x-y$, we get that $m\angle ECP=90$.
This proves that $\angle ECP$ is right.
(Spent way too long on this while talking to Blast_S1)
Muriatic
14.05.2017 23:32
We can solve this one by Vincent Huang bashing! Let $2\alpha$ be the angle measure of $\angle ACB$, and $\angle ACD = \beta$, so that $2 = \frac{AB}{CD} = \frac{AB}{AC}\cdot \frac{CD}{AC} = \frac{\sin 2\alpha}{\sin (2\alpha+\beta)}\cdot \frac{1}{\cos \beta}$. Let $x = e^{i\alpha}$ and $y = e^{i\beta}$ as usual, then $0<\alpha+\beta < \angle BCD <\pi$. Our condition is just $\left(x^2y - \frac{1}{x^2y}\right)\left(y + \frac 1y\right) = x^2-\frac 1{x^2}$, or $(xy)^4 = 1$. The only value of $xy$ that works is $i$, which solves the problem.
Stens
03.07.2017 01:45
Let $C'$ be the reflection of $C$ in $D$ , and let $X,Y$ be the intersections of the line through $B$ parallel with $CD$, and $AC,AC'$ respectively. Trivially, $AC'\parallel BC$ and hence $BCC'Y$ is a parallelogram. Moreover, $\triangle CAC'$ is isosceles because $D$ is both the midpoint and foot of the $A$-altitude on $CC'$. This means that \[ BC=YC'=XC, \]so $\triangle BCX$ is isosceles, and we are done. [asy][asy] import graph; size(6cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); real xmin=4.7,xmax=15.,ymin=-3.2,ymax=7.2; pen fueaev=rgb(0.9568627450980393,0.9176470588235294,0.8980392156862745), zzttqq=rgb(0.6,0.2,0.); pair A=(5.74995,-2.19), B=(12.5,-2.19), D=(11.544247289929165,3.7667910862532974), C=(13.963519579858335,1.4135135127813803), X=(10.40185117720903,-0.14908604788547042), Y=(7.661455420141661,2.516555146943839); filldraw(A--B--C--D--cycle,fueaev,linewidth(0.8)+zzttqq); draw((xmin,-0.9727212531090562*xmin+9.969015663863209)--(xmax,-0.9727212531090562*xmax+9.969015663863209),linewidth(0.8)); draw(C--A,linewidth(0.8)); draw(C--B,linewidth(0.8)); draw(D--A,linewidth(0.8)); draw(A--B,linewidth(0.8)+zzttqq); draw(B--C,linewidth(0.8)+zzttqq); draw(C--D,linewidth(0.8)+zzttqq); draw(D--A,linewidth(0.8)+zzttqq); draw((9.124974999999996,6.120068659725216)--A,linewidth(0.8)); draw((9.124974999999996,6.120068659725216)--D,linewidth(0.8)); dot(A); label("$A$",(5.0,-1.9488047913913684),NE*lsf); dot(B); label("$B$",(12.81435486084319,-1.995129896740937),NE*lsf); dot(D); label("$D$",(11.6330646744292,4.003971246028185),NE*lsf); dot(C); label("$C$",(14.180945468655455,1.5487406625010542),NE*lsf); dot(X); label("$X$",(10.1,0.08949984398964606),NE*lsf); dot((9.124974999999996,6.120068659725216)); label("$C'$",(9.0,6.343389066181395),NE*lsf); dot(Y); label("$Y$",(7.3,2.822681059614188),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]
hectorraul
24.10.2017 12:24
I think a have a new one, Let the angle bisector of $\angle ACB$ meet $AB$ at $K$, and let $AB\cap DC=L,\quad BC\cap AD = M$. Now, $BL=CL$, by Menelaus in triangles $ALD$ and $ABM$ we have $AD=DM$ which implies $AC=CM$, then $\frac{BC}{CM}=\frac{BC}{CA}=\frac{BK}{KA}$ and we deduce $KC\parallel AD$.
Projeto_Ramanujan
24.10.2017 20:42
Blast_S1 wrote:
First, extend $\overline{AB}$ and $\overline{DC}$ to meet at point $E$, and draw $\overline{AC}$
WLOG, let $CD=x, AB=2x, m\angle ACD=\phi, m\angle ABC=m\angle DCB=\theta$
$CLAIM:$ $EC=EB$
$PROOF:$
This is simple angle chasing, $m\angle ECB=m\angle EBC=180^\circ-\theta$
so $\triangle EBC$ is isosceles, and therefore $EB=EC$
we can now do a bit more angle chasing to get that $m\angle DAC=90^\circ-\phi$ and $m\angle ACB=\theta-\phi$
Consider $\triangle AEC$, we note that $\frac{AC}{EC}=\frac{\sin\angle AEC}{\sin\angle EAC}$ by a simple LOS
applying ratio lemma yields
$$\frac{AB}{BE}=\frac{2x}{BE}=\frac{AC\cdot\sin\angle ACB}{CE\cdot\sin\angle ECB}=\frac{\sin\angle AEC\cdot\sin\angle ACB}{\sin\angle EAC\cdot\sin\angle ECB}$$
similarly, for $\triangle AED$, we have $\frac{AD}{AE}=\frac{\sin\angle AEC}{\sin\angle ADC}$
however this time, since $m\angle ADE=90^\circ$, the above fraction simply reduces to $\sin\angle AEC$
now, using ratio lemma again, we have
$$\frac{DC}{CE}=\frac{x}{CE}=\frac{AD\sin\angle DAC}{AE\sin\angle EAC}=\frac{\sin\angle AEC\sin\angle DAC}{\sin\angle EAC}$$
finally, since $BE=CE$, therefore $\frac{2x}{BE}=2\cdot\frac{x}{CE}$, or
$$\frac{\sin\angle AEC\cdot\sin\angle ACB}{\sin\angle EAC\cdot\sin\angle ECB}=2\cdot\frac{\sin\angle AEC\sin\angle DAC}{\sin\angle EAC}\iff \sin\angle ACB=2\cdot\sin\angle ECB\cdot\sin\angle DAC$$or
$$\sin(\theta-\phi)=\sin(180^\circ-\theta)\sin(90^\circ-\phi)$$$$\sin\theta\cos\phi-\sin\phi\cos\theta=2\sin\theta\cos\phi$$$$0=\sin\theta\cos\phi+\sin\phi\cos\theta$$$$0=\sin(\theta+\phi)$$the only possible way for this to work is if $\theta+\phi=180^\circ\iff \phi=180^\circ-\theta$
this yields that $m\angle ACD=m\angle BCE=180^\circ-\theta$, so the angle bisector of $\angle BCA$ is clearly also the angle bisector of $\angle ECD$, which is a straight angle
and we're done bashng
Let $\angle{ACD}$ angle bisector meet $AB$ at $X$ Let $\alpha=\angle{ACB} \wedge \beta=\angle{ACD}$ Sine Law in $\triangle{ABC} \wedge \triangle{ACD}$ gives: $$\frac{2CD}{sin\alpha}=\frac{AC}{sin(\alpha+\beta)} \wedge AC cos\beta=CD$$hence $$2sin(\alpha+\beta)cos\beta=sin\alpha$$using product-to-sum tranformation, we discover $sin(\alpha+2\beta)=0\Rightarrow \alpha+2\beta=180$, hence proved, since $\angle{DCX}=\beta+\alpha/2=90^{\circ}$.
GRCMIRACLES
20.01.2018 11:41
Apply sine rule in triangle ACD we get AC=a/cosx where x=angle ACD,a=CD now sine rule in triangle ABC =>a/{cosxsiny}=2a/sin(y-x) where y=angle BCD =>By solving the above equation,we get sin(y-x)=2{cosx.siny},by using transformations we get x+y=180 =>we should prove that x+{(y-x)/2}=90
Blast_S1
05.02.2018 21:04
Blast_S1 wrote:
First, extend $\overline{AB}$ and $\overline{DC}$ to meet at point $E$, and draw $\overline{AC}$
WLOG, let $CD=x, AB=2x, m\angle ACD=\phi, m\angle ABC=m\angle DCB=\theta$
$CLAIM:$ $EC=EB$
$PROOF:$
This is simple angle chasing, $m\angle ECB=m\angle EBC=180^\circ-\theta$
so $\triangle EBC$ is isosceles, and therefore $EB=EC$
we can now do a bit more angle chasing to get that $m\angle DAC=90^\circ-\phi$ and $m\angle ACB=\theta-\phi$
Consider $\triangle AEC$, we note that $\frac{AC}{EC}=\frac{\sin\angle AEC}{\sin\angle EAC}$ by a simple LOS
applying ratio lemma yields
$$\frac{AB}{BE}=\frac{2x}{BE}=\frac{AC\cdot\sin\angle ACB}{CE\cdot\sin\angle ECB}=\frac{\sin\angle AEC\cdot\sin\angle ACB}{\sin\angle EAC\cdot\sin\angle ECB}$$
similarly, for $\triangle AED$, we have $\frac{AD}{AE}=\frac{\sin\angle AEC}{\sin\angle ADC}$
however this time, since $m\angle ADE=90^\circ$, the above fraction simply reduces to $\sin\angle AEC$
now, using ratio lemma again, we have
$$\frac{DC}{CE}=\frac{x}{CE}=\frac{AD\sin\angle DAC}{AE\sin\angle EAC}=\frac{\sin\angle AEC\sin\angle DAC}{\sin\angle EAC}$$
finally, since $BE=CE$, therefore $\frac{2x}{BE}=2\cdot\frac{x}{CE}$, or
$$\frac{\sin\angle AEC\cdot\sin\angle ACB}{\sin\angle EAC\cdot\sin\angle ECB}=2\cdot\frac{\sin\angle AEC\sin\angle DAC}{\sin\angle EAC}\iff \sin\angle ACB=2\cdot\sin\angle ECB\cdot\sin\angle DAC$$or
$$\sin(\theta-\phi)=\sin(180^\circ-\theta)\sin(90^\circ-\phi)$$$$\sin\theta\cos\phi-\sin\phi\cos\theta=2\sin\theta\cos\phi$$$$0=\sin\theta\cos\phi+\sin\phi\cos\theta$$$$0=\sin(\theta+\phi)$$the only possible way for this to work is if $\theta+\phi=180^\circ\iff \phi=180^\circ-\theta$
this yields that $m\angle ACD=m\angle BCE=180^\circ-\theta$, so the angle bisector of $\angle BCA$ is clearly also the angle bisector of $\angle ECD$, which is a straight angle
and we're done bashng
[asy][asy]
size(5cm);
pair A=(0,120), B=(40,0), C=(160,0), D=(180,60), E=(200,120);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
draw(A--B--C--D--cycle, linewidth(0.5));
draw(A--C, linewidth(0.5));
draw(A--E--D, linewidth(0.5)+dashed);
label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, (0.7,0));
label("$E$", E, NE);
[/asy][/asy]
Extend $\overline{CD}$ to $E$ such that $CE=2CD$. Evidently, $ABCE$ is an isosceles trapezoid and $\triangle ACE$ is isosceles. This means that
$$\angle ECA=\angle CEA=90^\circ-\frac{\angle CAE}{2}=90^\circ-\frac{\angle ACB}{2},$$which immediately implies the desired conclusion.
DimPlak
30.04.2018 12:51
How can we make the construction of the figure with ruler and compass?
WolfusA
30.04.2018 18:26
Stens wrote: Let $C'$ be the reflection of $C$ in $D$ , and let $X,Y$ be the intersections of the line through $B$ parallel with $CD$, and $AC,AC'$ respectively. Trivially, $AC'\parallel BC$ and hence $BCC'Y$ is a parallelogram. Moreover, $\triangle CAC'$ is isosceles because $D$ is both the midpoint and foot of the $A$-altitude on $CC'$. This means that \[ BC=YC'=XC, \]so $\triangle BCX$ is isosceles, and we are done. [asy][asy] import graph; size(6cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); real xmin=4.7,xmax=15.,ymin=-3.2,ymax=7.2; pen fueaev=rgb(0.9568627450980393,0.9176470588235294,0.8980392156862745), zzttqq=rgb(0.6,0.2,0.); pair A=(5.74995,-2.19), B=(12.5,-2.19), D=(11.544247289929165,3.7667910862532974), C=(13.963519579858335,1.4135135127813803), X=(10.40185117720903,-0.14908604788547042), Y=(7.661455420141661,2.516555146943839); filldraw(A--B--C--D--cycle,fueaev,linewidth(0.8)+zzttqq); draw((xmin,-0.9727212531090562*xmin+9.969015663863209)--(xmax,-0.9727212531090562*xmax+9.969015663863209),linewidth(0.8)); draw(C--A,linewidth(0.8)); draw(C--B,linewidth(0.8)); draw(D--A,linewidth(0.8)); draw(A--B,linewidth(0.8)+zzttqq); draw(B--C,linewidth(0.8)+zzttqq); draw(C--D,linewidth(0.8)+zzttqq); draw(D--A,linewidth(0.8)+zzttqq); draw((9.124974999999996,6.120068659725216)--A,linewidth(0.8)); draw((9.124974999999996,6.120068659725216)--D,linewidth(0.8)); dot(A); label("$A$",(5.0,-1.9488047913913684),NE*lsf); dot(B); label("$B$",(12.81435486084319,-1.995129896740937),NE*lsf); dot(D); label("$D$",(11.6330646744292,4.003971246028185),NE*lsf); dot(C); label("$C$",(14.180945468655455,1.5487406625010542),NE*lsf); dot(X); label("$X$",(10.1,0.08949984398964606),NE*lsf); dot((9.124974999999996,6.120068659725216)); label("$C'$",(9.0,6.343389066181395),NE*lsf); dot(Y); label("$Y$",(7.3,2.822681059614188),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] $BCC'Y$ is a parallelogram? You didn't prove $AC'=BC$.
WolfusA
02.05.2018 20:08
Assumptions like in first post. Prove that bisector of angle $ABC$ isn't parralel to line $CD$
dikhendzab
19.04.2020 12:56
Let bisector of $\angle ACB$ intersects side $AB$ at point $X$, let sides $BC$ and $AD$ intersect at\ point $E$ and let $F$ be the midpoint of side $AB$. Now, quadrilateral $FBCD$ is isosceles trapezoid, since $FB=CD$ and $\angle FBC=\angle BCD$, so $FD \parallel BC$ and $FD$ is midline of triangle $ABE$. Triangle $ACE$ is isosceles since $CD$ bisects $AE$ and it is also an altitude, so it also bisects $\angle ACE$. Now, we have: $\angle DCX=\angle DCA+\angle ACX=\frac{\angle ACE+\angle ACB}{2}=\frac{180^\circ}{2}=90^\circ$.
bin_sherlo
17.09.2023 14:54
Let the perpendicular to $CD$ at $C$ intersect $AB$ at $M$. We'll prove that $CM$ bisects $\angle ACB$. $AB \cap CD=K,AD \cap BC=L$ By Menelaus \[\frac{LD}{LA}.2=\frac{LD}{LA}.\frac{BA}{BK}.\frac{CK}{CD}=1\]So $LD=DA$ which gives $90-\angle MCB=\angle LCD=\angle DCA=90-\angle ACM$ $\implies \angle MCB=\angle ACM$ as desired.
