AD=BE implies ABC right

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

6874 posts

#1 h Apr 10, 2013, 3:12 PM • 16 Y

Y by Davi-8191, thedoge, microsoft_office_word, ImSh95, son7, Jc426, HWenslawski, Amiralishafiei, HamstPan38825, TheHawk, Adventure10, Mango247, ItsBesi, MS_asdfgzxcvb, NicoN9, Yiyj1

The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$ . The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$ . Prove that, if $AD=BE$ , then the triangle $ABC$ is right-angled.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

leader

339 posts

leader

#2 h Apr 10, 2013, 3:24 PM • 4 Y

Y by thedoge, HWenslawski, Adventure10, Mango247

let $X$ be symmetric point of $A$ wrt $C$ than $ABXD$ is a parallelogram and $\angle CAD=\angle BXE$ and $BX=AD=BE$
so let $M$ be the midpoint of $AE$ since $\angle BEM=\angle BXE=\angle CAD$ , $CA=AE/2=ME$ and $AD=BE$ we have $BEM\cong CAD$ so $BM=CD=CB$ so since $BMC$ is $B$ -isosceles and $A$ is the midpoint of $CM$ then $\angle BAC=90$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Number1

355 posts

Number1

#3 h Apr 10, 2013, 4:57 PM • 11 Y

Y by v_Enhance, leader, MSTang, biomathematics, thedoge, govind7701, Illuzion, Not_real_name, egxa, Adventure10, Mango247

Let $M$ be midpoint for $BE$ .

We see that $A$ is gravty center for triangle $BDE$ . Then $M$ is circumcenter for triangle $ABE$ thus $\angle BAC = \pi/2$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

v_Enhance

6874 posts

v_Enhance

#4 h Apr 11, 2013, 3:22 AM • 7 Y

Y by thczarif, guadasola, ImSh95, Jc426, HamstPan38825, Adventure10, Yiyj1

Nice solutions! Unfortunately this is pretty easy to bash with the cosine law, which is probably why it's a problem 1. Indeed, applying the law of cosines to $\triangle DCA$ and $\triangle ECB$ and solving for $\cos \angle ACB$ directly gives $\cos \angle ACB = \tfrac{CA}{CB}$ , implying the conclusion.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Particle

179 posts

Particle

#5 h Apr 11, 2013, 7:52 AM • 2 Y

Y by Adventure10 and 1 other user

Suppose $M$ is the mid-point of $AB$ and $N$ is the mid-point of $AE$ . So we see $MN=\frac 1 2 BE=\frac 1 2 AD=MC$ . Also $NA=\frac 1 2 AE=AC$ . So $\triangle MAN\cong MAC\implies \angle MAN=\angle MAC=\pi /2$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

yunxiu

571 posts

yunxiu

#6 h Apr 11, 2013, 8:41 AM • 1 Y

Y by Adventure10

Denote $AC = x$ , $BC = y$ , $BE = AD = z$ , By $Stewart$ theorem ${x^2} + {y^2} = \frac{1}{2}A{B^2} + \frac{1}{2}{z^2}$ ①, and $A{B^2} + 2{x^2} = \frac{2}{3}{y^2} + \frac{1}{3}{z^2}$ ②.
① $\times 2 -$ ② $\times 3$ we have ${y^2} = {x^2} + A{B^2}$ , so $\angle BAC = 90^\circ$ .

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Orin

22 posts

Orin

#7 h Apr 12, 2013, 1:51 PM • 4 Y

Y by E2lmontes, Adventure10, Mango247, and 1 other user

Let $F$ be the reflection of $A$ in $C$ . So $AE=AF$ and $ABFD$ is a parallelogram
which implies $BE=AD=BF$ . So $\triangle ABE \cong \triangle ABF \Rightarrow \angle BAC=90^{\circ}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

XtremeBanana5

73 posts

XtremeBanana5

#8 h Apr 14, 2013, 3:29 PM • 2 Y

Y by Adventure10, Mango247

LoC on $\triangle ABE$ : $BE^2 = 4b^2 + c^2 + 4bc \cos A$
LoC on $\triangle ABC$ : $2a^2 = 2b^2 + 2c^2 - 4bc \cos A$
Adding and solving for $BE$ gives $BE^2 = 6b^2 + 3c^2 - 2a^2$

LoC on $\triangle ACD$ : $AD^2 = a^2 + b^2 + 2ab \cos C$
LoC on $\triangle ABC$ : $c^2 = a^2 + b^2 - 2ab \cos C$
Solving for $AD$ gives $AD^2 = 2a^2 + 2b^2 - c^2$

Setting these two equal we get
$6b^2 + 3c^2 - 2a^2 = 2a^2 + 2b^2 - c^2$
or $4b^2 + 4c^2 = 4a^2$ so $\triangle ABC$ is right

This post has been edited 1 time. Last edited by v_Enhance, Apr 14, 2013, 4:18 PM
Reason: Use \cos.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Abu3gab

8 posts

Abu3gab

#9 h May 13, 2013, 4:13 PM • 2 Y

Y by Adventure10, Mango247

http://www.artofproblemsolving.com/Forum/download/file.php?mode=view&id=43658&

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

aZpElr68Cb51U51qy9OM

1600 posts

aZpElr68Cb51U51qy9OM

#10 h May 14, 2013, 8:24 PM • 6 Y

Y by v_Enhance, GoJensenOrGoHome, AlastorMoody, myh2910, Adventure10, Mango247

We use barycentric coordinates. Let $A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)$ , and let $BC = a, CA = b, AB = c$ .

Since $\frac{BC}{CD} = 1$ and $\frac{CA}{AE} = \frac{1}{2}$ , we have that $D = (0, -1, 2)$ and $E = (3, 0, -2)$ . By the Distance Formula, the lengths of displacement vectors $\overrightarrow{AD} = (-1, -1, 2)$ and $\overrightarrow{BE} = (3, -1, -2)$ can be found:
$|AD|^2 = -2a^2-2b^2+c^2$ $|BE|^2 = 2a^2 - 6b^2 - 3c^2$
The condition that $AD = BE$ implies that $4a^2 + 4b^2 = 4c^2$ , so $a^2+b^2=c^2$ ; therefore, $\triangle ABC$ is right.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

djmathman

7938 posts

djmathman

#11 h May 16, 2013, 3:36 AM • 3 Y

Y by Jc426, Adventure10, Mango247

Let $BC=a$ , $CA=b$ , $AB=c$ . In addition, use $A$ , $B$ , $C$ as shorthands for $\angle BAC$ , $\angle ABC$ , $\angle ACB$ respectively. By LoC on $\triangle ACD$ , we get that $AD^2=b^2+a^2-2ab\cos\angle ACD=b^2+a^2+2ab\cos C.$ By Law of Cosines on $\triangle ABE$ , we get that $BE^2=(2b)^2+c^2-2(2b)(c)\cos\angle EAB=4b^2+c^2+4bc\cos A.$ Combining these two gives $\begin{align*}a^2+b^2+2ab\cos\angle C &= 4b^2 + c^2 + 4bc\cos A= 4b^2 + (a^2+b^2-2ab\cos C) + 4bc\cos A\\ 4ab\cos C &= 4b^2 + 4bc\cos A\\ b &= a\cos C-c\cos A,\end{align*}$ which (I believe) holds true iff $A=90^\circ$ . $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

NewAlbionAcademy

910 posts

NewAlbionAcademy

#12 h May 29, 2013, 12:03 AM • 2 Y

Y by Adventure10, balllightning37

Solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

srinidhi321

132 posts

srinidhi321

#13 h May 29, 2013, 1:52 PM • 3 Y

Y by ItsBesi, Adventure10, Mango247

let $AD$ meet $BE$ at $P$ and $X$ be the mid point of $AE$
Now, $A$ is the centroid of $\triangle EBD$
So $PA=\frac { AD }{ 2 }$
And $P$ is the mid point of $EB$ So $PA=\frac { AD }{ 2 }$
So $PE=PA$ and hence $\triangle PAE$ is isosceles .
So, $PX\bot AE$ but $PX\parallel BA$
So. $BA\bot AC$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

AllonsyAllonso

148 posts

AllonsyAllonso

#14 h Jul 5, 2013, 3:28 PM • 6 Y

Y by Mathematist07, Aryan-23, lambda5, Adventure10, Mango247, and 1 other user

Let $ABXD$ parallelogram. $AX$ intersects $BD$ at $C$ . Then $EB=AD=BX$ and $EA=AX$ because of $ABXD$ parallelogram and $\frac{EA}{2}=AC=BX$ . So $A$ is the midpoint of the $EX$ and $EBX$ isosceles triangle then $BA\bot AC$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Mathematist07

1 post

Mathematist07

#15 h Jul 14, 2013, 1:33 PM • 1 Y

Y by Adventure10

AllonsyAllonso wrote:

Let $ABXD$ parallelogram. $AX$ intersects $BD$ at $C$ . Then $EB=AD=BX$ and $EA=AX$ because of $ABXD$ parallelogram and $\frac{EA}{2}=AC=BX$ . So $A$ is the midpoint of the $EX$ and $EBX$ isosceles triangle then $BA\bot AC$ .

Very nice solution thanks

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

AshAuktober

993 posts

AshAuktober

#104 h Mar 7, 2024, 4:55 PM

Y by

Apply the cosine rule on triangles $ABE$ and $ACD$ , to get $4b^2 + c^2 + 4bc\cos A = a^2 + b^2 + 2ab\cos C.$ From cosine rule on triangle $ABC$ , $c^2 = a^2 + b^2 - 2ab\cos C.$ Subbing in, $4b^2 + 4bc\cos A = 4ab\cos C \implies b + c\cos A = a\cos C.$ But it is well known that $b = c\cos A + a\cos C$ . Adding the equations, $c\cos A = 0 \implies \cos A = 0 \implies \angle A = \frac{\pi}{2}$ $\square$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

blueberryfaygo_55

340 posts

blueberryfaygo_55

#105 h Apr 3, 2024, 7:27 PM

Y by

Let $X = DA \cap BE$ . By Menelaus' Theorem on the transversal $\overline{XAS}$ in $\Delta BEC$ , we have $\dfrac{BX}{XE} \cdot \dfrac{EA}{AC} \cdot \dfrac{CD}{DB} = 1$ which leads to $\dfrac{BX}{XE} = 1$ , or $BX=XE$ . Since $C$ is also the midpoint of $AD$ , it follows that $A$ is the centroid of $\Delta BED$ , which implies that $AX = \dfrac{AD}{2}=\dfrac{BE}{2}$ This is equivalent to $\Delta BAE$ being right-angled at $A$ , or $\angle BAE = \angle BAC = 90^{\circ}$ , as desired. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

BestAOPS

707 posts

BestAOPS

#106 h Apr 28, 2024, 6:24 PM

Y by

Let $a$ , $b$ , $c$ denote the side lengths of $ABC$ in their usual way. We can compute
$\begin{align*} AD^2 &= c^2 + 4a^2 - 4ac \cos B \\ BE^2 &= c^2 + 4b^2 + 4bc \cos A. \end{align*}$ (The $+$ is not a mistake in the second line there!) Equating the two, we get $a^2 - ac \cos B = b^2 + bc \cos A$ . Using the Law of Cosines but solving for angles, we get
$\begin{align*} \cos B &= \frac{a^2 + c^2 - b^2}{2ac} \\ \cos A &= \frac{b^2 + c^2 - a^2}{2bc}. \end{align*}$ Plugging these back in, we can simplify to get $a^2 = b^2 + c^2$ . Thus, triangle $ABC$ is right-angled.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

blueberryfaygo_55

340 posts

blueberryfaygo_55

#107 h May 17, 2024, 3:27 PM

Y by

Let $BC = CD = a$ , $AC = b$ , $BA = c$ , $EA = 2b$ , and $BE = AD = d$ .

By the Law of Cosines in $\Delta ACD$ , we have $d^2 = a^2 + b^2 - 2ab \cos ACD.$ By the Law of Cosines in $\Delta EBC$ , we have $d^2 = a^2 + 9b^2 - 6ab \cdot ACB.$ Equating the expressions above yields $\begin{align*} a^2 + b^2 - 2ab \cos ACD &= a^2 + 9b^2 - 6ab \cos ACB \\ 8b^2 - 6ab \cos ACB = -2ab \cos ACD \\ 8b^2 = 8ab \cos ACB \\ \cos ACB = \dfrac ba . \end{align*}$ Thus, applying the Law of Cosines in $\Delta ABC$ gives $\begin{align*} c^2 &= a^2 + b^2 - 2ab \cos ACB \\ c^2 &= a^2 + b^2 - 2b^2 \\ c^2 &= a^2 - b^2 \end{align*}$ and the desired condition follows by the Converse of Pythagorean Theorem. $\blacksquare$

This post has been edited 1 time. Last edited by blueberryfaygo_55, May 17, 2024, 3:28 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Tellocan

35 posts

Tellocan

#108 h May 31, 2024, 2:07 PM

Y by

v_Enhance wrote:

The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$ . The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$ . Prove that, if $AD=BE$ , then the triangle $ABC$ is right-angled.

Posting for storage...
Let the midpoint of $AE$ be $F$ . Then, note that $EF=AC=x$ . Bu using the Sine Law on $\triangle EBC$ and $\triangle ACD$ , we get that $\angle BEC=\angle CAD= a$ . So, $\triangle FEB$ is congruent to $\triangle CAD$ . So $BF=CD=CB$ . As $AF=AC$ ,we get that $\angle BAC=90$ , as desired.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Scilyse

387 posts

Scilyse

#109 h Sep 28, 2024, 3:25 PM

Y by

Note that $A$ is the centroid of triangle $EBD$ . Thus let $M = AD \cap BE$ , noting that $BE = AD = 2AM$ and that as $M$ is the midpoint of $\overline{BE}$ , we have $\angle BAE = \angle BAC = 90^\circ$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

redred123

32 posts

redred123

#113 h Oct 7, 2024, 9:20 PM

Y by

let $M$ be the midpoint of $BC$ and let $B(-1,0)$ and $C(1,0)$ and $A(m,n)$ then it is sufficient to prove that $m^2+n^2=1$ but from the conditions of the exercise we have $E(3m-2,3n)$ and $D(3,0)$ and from the exercice condition we have $(3m-1)^2+(3n)^2=(m-3)^2+n^2$ which yields to the fact that $\triangle ABC$ is right angled at $A$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Likeminded2017

391 posts

Likeminded2017

#114 h Dec 26, 2024, 8:37 AM

Y by

Let the line through $A$ parallel to $BC$ intersect $BE$ and let $DA$ intersect $BE$ at $F.$ Projecting through $A,$ we have $-1=(B,D;C,P_{\infty})=(B,F;E,G)$ so
$1=\dfrac{\frac{BE}{FE}}{\frac{BG}{FG}}=\dfrac{FG}{FE} \cdot 3,$ so $FE=\dfrac{2}{3} \cdot \dfrac{3}{4}z=\dfrac{1}{2}z$ and $FB=\dfrac{1}{2}z.$ Applying Menelaus on $\triangle BDF$ with line $CE$ we find $FA=\dfrac{1}{2}z$ so $FA=FE=FB$ and thus $\triangle AEB$ is right, meaning $\angle BAC=90^\circ.$

Update from MaxSze:

It is much easier to deduce $FE=FB$ using centroid properties by constructing $\triangle DEB$ with centroid $A.$ The solution then follows naturally.

This post has been edited 2 times. Last edited by Likeminded2017, Dec 26, 2024, 8:43 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Maximilian113

550 posts

Maximilian113

#115 h Jan 2, 2025, 11:44 PM

Y by

Let $x=BC, y=AC$ then by the Law of Cosines $9y^2+x^2-6xy\cos C =BE^2=AD^2= x^2+y^2+2xy\cos C \implies 8y^2=8xy\cos C \implies \cos C = \frac{y}{x}.$ Now, draw the perpendicular to $AC$ at $A,$ and suppose that it intersects $BC$ at $P.$ Then $\cos C = \frac{y}{PC} \implies PC=x,$ therefore $\angle A = 90^\circ$ so we are done. QED

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Siddharthmaybe

106 posts

Siddharthmaybe

#116 h Jan 3, 2025, 10:31 AM

Y by

Let BC = CD = a, AC = b, EA = 2b, AD = BE = c

By Appolonius Theorem, on triangle ABD we get AB^2 = x^2 = 2(a^2 + b^2) - c^2

By law of cosines on triangle ABC and BAE with respect to angles CAB and BAE = 180 - CAB (and comparing the equations) we get :-

a^2 + 3b^2 - c^2 = 0
or 2a^2 + 2b^2 - c^2 = a^2 - b^2
or x^2 + b^2 = a^2 and note that this satisfies the condition for right angled triangle ABC by the Pythagorean theorem and so we are done

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

clarkculus

224 posts

clarkculus

#117 h Feb 9, 2025, 10:56 PM • 1 Y

Y by centslordm

Let $AC=x$ , $AE=2x$ , $BC=CD=y$ , $BE=AD=z$ and $AB=d$ . By Stewart's on $\triangle BEC$ , $6x^2+3d^2=z^2+2y^2$ . By Stewart's on $\triangle BAD$ , $2y^2+2x^2=d^2+z^2$ . Eliminating $z^2$ , we have $x^2+d^2=y^2$ , implying $\triangle ABC$ has a right angle at $A$ , as desired.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Mathgloggers

67 posts

Mathgloggers

#118 h Mar 18, 2025, 1:59 PM

Y by

math_pi_rate wrote:

Another way to end: Note that $A$ is the centroid of $\triangle BDE$ . Then $BE^2=AD^2=\frac{4}{9} \times \frac{1}{4}(2DE^2+2BD^2-BE^2) \Rightarrow 10BE^2=2DE^2+2BD^2$ Thus, $AB^2+AC^2=\frac{4}{9} \times \frac{1}{4}(2BE^2+2BD^2-DE^2)+\frac{1}{9} \times \frac{1}{4}(2DE^2+2BE^2-BD^2)=\frac{1}{36}(10BE^2+7BD^2-2DE^2)=\frac{9BD^2}{36}=BC^2$ This gives that $\triangle ABC$ is right-angled at $\angle BAC$ .

Remark (Just for Indians

)- After recognising that $A$ is the centroid of $\triangle BDE$ , this problem is basically in this year's KVPY SA paper

.

ALSO FOR IOQM 2023 P7 I GUESS

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Marcus_Zhang

979 posts

Marcus_Zhang

#119 h Apr 6, 2025, 2:21 AM

Y by

trig-bashed

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

eg4334

631 posts

eg4334

#120 h Apr 8, 2025, 2:19 AM

Y by

If $AC = x, BC = CD = y, \angle ACB = \theta$ , then by LOC on $\triangle BCE$ and $\triangle ACD$ we have $9x^2+y^2-6xy \cos{\theta} = x^2+y^2+2xy \cos{\theta} \implies \cos{\theta} = \frac{x}{y}$ . This obviously implies the desired by trig intuition, or by the law of cosines again which reduces to pythagorean theorem.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

LeYohan

41 posts

LeYohan

#121 h Apr 13, 2025, 2:02 PM

Y by

Focusing on $\triangle BED$ , notice that $DA$ divides $CE$ in a $2 : 1$ ratio, and since $CE$ bisects $BD$ we can see that $A$ is the centroid of $\triangle BED$ .

Let $X = BE \cap AD$ , and because $AD = BE$ , then $AX = 2AD = 2BE \implies XE = XB = XA$ , so $\triangle ABE$ is right-angle on $A$ , as desired. $\square$

Z K Y