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 $\angle BAC=90$

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

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.

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 $\triangle MAN\cong MAC\implies \angle MAN=\angle MAC=\pi /2$.

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 $.

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

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

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.

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 $A=90^\circ$. $\blacksquare$

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. $BA\bot AC$

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 $.

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

Let $ F $ be the point on line $ AB $ such that $ DF \parallel CA $. $ DF = 2AC $ by similar triangles, and $ AE = 2AC $. Note also that $ AB = AF $. Thus, $ FDA \cong AEB $. By the Law of Cosines, $ \cos(DFA) = \cos(180 - DFA) \Rightarrow DFA = CAB = 90 ^\circ $.

[asy][asy] unitsize(1.5cm); pointpen=black; pathpen=black; pair A = Drawing("A", (0,0), dir(180)); pair B = Drawing("B", (1.6,0), dir(0)); pair C = Drawing("C", (0,1), dir(45)); draw(A--B--C--cycle); pair D = Drawing("D", (-1.6,2), dir(90)); pair E = Drawing("E", (0,-2)); draw(B--D); pair F = Drawing("F", (0,2), dir(90)); draw(E--F); draw(E--B--F--D--A); [/asy][/asy] Let $F$ be such that $ABFD$ is a parallelogram. Then since $BC=CD$, $C$ is the midpoint of $AF$. $\therefore AF = 2AC = AE$, so $BA$ bisects $EF$. But $BF = AD = BE$, so $\triangle BEF$ is isosceles. It follows $BA \perp EF$, and $\triangle ABC$ is right-angled.

CD=BC implies that EC is the barycentric line of the triangle EBD, and obviously because EA:AC=2:1, a is the barycenter of that triangle. Hence, AD meets EB in its center, lets name it F, and 2AF=AD=EB, hence AF=EF=BF, so the angle EAB is right, so the angle CAB is also right.

djmathman wrote: Combining these two gives \begin{align*}b &= a\cos C-c\cos A,\end{align*} which (I believe) holds true iff $A=90^\circ$. $\blacksquare$ Wait goddangit this is probably the sketchiest thing I've ever written on this site Fortunately this claim is easy to prove. The 'if' direction is trivial. To work the 'only if' direction, let $D$ be the projection of $B$ onto $AC$. Then $a\cos C-c\cos A=DC-DA$. If this equals $b$, then we need both $DC-DA=b$ and $DC+DA=b$, so $DA=0$, which implies $D\equiv A$.

Number1 wrote: Let $M$ be midpoint for $BE$. We see that $A$ is gravty center for triangle $BDE$. Then $M$ is circumcenter for triangle $ABE$ thus $\angle BAC = \pi/2$. Why is $M$ the circumcenter of $\triangle ABE?$ I don't see this following from the fact that $A$ is the centroid of $BDE.$

Solved over a year ago but writing up now cuz IDK why. :3 As $C$ is the midpoint of $BD$ in $\triangle EBD$ and $EA:AC = 2:1$, thus $A$ is the centroid. Let $X = DA\cap EB$. Then $DA:AX = 2:1\implies AX = \dfrac{DA}{2}$. Also, as $DA$ is the median, we get that $X$ is the midpoint of $EB\implies XE = XB\implies SB=\dfrac{EB}{2}=\dfrac{AD}{2}=XA$. Thus $XB=XE=XA\implies \angle EAB = 90^\circ \implies \angle BAC = 90^\circ$.

It is given that $BC=CD$ and $2CA=AE$. Joining $EB$ and $ED$. In triangle $\Delta EDB$, we get that $EC$ is a median. As $CA:AE=2:1$, we conclude that $A$ is the centroid of $\Delta EDB$. Let $BA \cap ED = F$, $DA \cap EB= G$. We get that $GE=GB$. Also, it is given that $AD=BE$. $AD:GA=2:1 \Rightarrow GA=AD/2$ Also, $AD=BE=EB/2=AD/2 \Rightarrow GA=EG=BG$ $\Rightarrow G$ is the circumcentre of $\Delta AEB$ $\Rightarrow \angle EAB= 90^{\circ}$ $\Rightarrow \angle BAC= 90^{\circ}$ $\framebox{Q.E.D}$.

Claim: $A$ is the centroid of $\Delta EBD$ Proof: $AE:AC=2:1$. Moreover, $BC=CD.$ So, it is quite obvious that $A$ is the centroid of $\Delta EBD$ as it is the only point in the median which divides the median in such a ratio. We assume that $AD$ intersects $BE$ at $F$. Now, $AD=BE \implies \frac {AD}{2}= \frac {BE}{2} \implies AF=BF=FE$ (as $AD:AF=2:1$) So, $\angle EAB = \frac{\pi}{2} \implies \boxed {\angle BAC = \frac{\pi}{2}}$

I knew many bary solution already existed, but complex is annoying on this problem so we still use bary. Take $ABC$ as the reference triangle. Use bary. $D=(0,1,-2)$ and $E=(3,0,2)$. Now use the distance formula to get \[|AD|^2=2a^2+2b^2-c^2=-2a^2+6b^2+3c^2\]Which simplifies to $4a^2+4b^2=4c^2$, so $ABC$ is right triangle.

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$

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. $\blacksquare$

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.

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. $\blacksquare$

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.

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 $\angle BAE = \angle BAC = 90^\circ$.

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 $A$

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.

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

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