Subfield: Geometry


Question: The side $B C$ of the triangle $A B C$ is extended beyond $C$ to $D$ so that $C D=B C$. The side $C A$ is extended beyond $A$ to $E$ so that $A E=2 C A$.

Prove that if $A D=B E$, then the triangle $A B C$ is right-angled.

Solution:
Define $F$ so that $A B F D$ is a parallelogram. Then $E, A, C, F$ are collinear (as diagonals of a parallelogram bisect each other) and $B F=A D=B E$. Further, $A$ is the midpoint of $E F$, since $A F=2 A C$, and thus $A B$ is an altitude of the isosceles triangle $E B F$ with apex $B$. Therefore $A B \perp A C$.

<img_3700>
Notice that $A$ is the centroid of triangle $B D E$, since $C$ is the midpoint of $[B D]$ and $A E=2 C A$. Let $M$ be the midpoint of $[B E]$. Then $M, A, D$ lie on a line, and further, $A M=\frac{1}{2} A D=\frac{1}{2} B E$. This implies that $\angle E A B=90^{\circ}$.



<img_3193>
Let $P$ be the midpoint $[A E]$. Since $C$ is the midpoint of $[B D]$, and, moreover, $A C=E P$, we have

$$
[A C D]=[A B C]=[E B P] .
$$

But $A D=B E$, and, as mentioned previously, $A C=E P$, so this implies that

$$
\angle B E P=\angle C A D \text { or } \angle B E P=180^{\circ}-\angle C A D \text {. }
$$

But $\angle B E P<\angle B A C$ and $\angle B A C+\angle C A D=\angle B A D<180^{\circ}$, so we must be in the first case, i.e. $\angle B E P=\angle C A D$. It follows that triangles $B E P$ and $D A C$ are congruent, and thus $\angle B P A=\angle A C B$. But $A P=A C$, so $B A$ is a median of the isosceles triangle $B C P$. Thus $A B \perp P C$, completing the proof.

<img_3776>
Write $\beta=\angle E C B$, and let $x=A C, y=B C=C D, z=B E=A D$. Notice that $E C=3 x$. Then, using the cosine theorem,

$$
\begin{array}{ll}
z^{2}=x^{2}+y^{2}+2 x y \cos \beta & \text { in triangle } A C D \\
z^{2}=9 x^{2}+y^{2}-6 x y \cos \beta & \text { in triangle } B C E
\end{array}
$$

Hence $4 z^{2}=12 x^{2}+4 y^{2}$ or $z^{2}-y^{2}=3 x^{2}$. Let $H$ be the foot of the perpendicular through $B$ to $A C$, and write $h=B H$. Then

$$
y^{2}-h^{2}=C H^{2}, z^{2}-h^{2}=E H^{2} .
$$



Hence $z^{2}-y^{2}=E H^{2}-C H^{2}$. Substituting from the above,

$$
E H^{2}-C H^{2}=3 x^{2}=E A^{2}-C A^{2}
$$

Thus $H=A$, and hence the triangle $A B C$ is right-angled at $A$.
Writing $a=B C, b=C A, c=A B$, we have

$$
\begin{array}{rlrl}
a^{2} & =b^{2}+c^{2}-2 b c \cos \angle A & & \text { in triangle } A B C ;\\
c^{2} & =a^{2}+b^{2}-2 a b \cos \angle C & & \text { in triangle } A B C ;\\
E B^{2} & =4 b^{2}+c^{2}+4 b c \cos \angle A  & & \text { in triangle } A E B ;\\
A D^{2} & =a^{2}+b^{2}+2 a b \cos \angle C \quad & & \text { in triangle } A C D ;
\end{array}
$$

Thus

$$
\begin{aligned}
6 b^{2}+3 c^{2}-2 a^{2} & =4 b^{2}+c^{2}+4 b c \cos \angle A=E B^{2}=A D^{2} \\
& =a^{2}+b^{2}+2 a b \cos \angle C=2 a^{2}+2 b^{2}-c^{2}
\end{aligned}
$$

which gives $a^{2}=b^{2}+c^{2}$. Therefore $\angle B A C$ is a right angle by the converse of the theorem of Pythagoras.
Let $\overrightarrow{A C}=\vec{x}$ and $\overrightarrow{A B}=\vec{y}$. Now $\overrightarrow{A D}=2 \vec{x}-\vec{y}$ and $\overrightarrow{E B}=2 \vec{x}+\vec{y}$. Then

$$
\overrightarrow{B E} \cdot \overrightarrow{B E}=\overrightarrow{A D} \cdot \overrightarrow{A D} \quad \Longleftrightarrow \quad(2 \vec{x}+\vec{y})^{2}=(2 \vec{x}-\vec{y})^{2} \Longleftrightarrow \vec{x} \cdot \vec{y}=0
$$

and thus $A C \perp A B$, whence triangle $A B C$ is right-angled at $A$.
Let $a, b, c, d$, $e$ denote the complex co-ordinates of the points $A, B, C, D$, $E$ and take the unit circle to be the circumcircle of $A B C$. We have

$$
d=b+2(c-b)=2 c-b \quad \text { and } \quad e=c+3(a-c)=3 a-2 c \text {. }
$$

Thus $b-e=(d-a)+2(b-a)$, and hence

$$
\begin{aligned}
B E=A D & \Longleftrightarrow(b-e)(\overline{b-e})=(d-a)(\overline{d-a}) \\
& \Longleftrightarrow 2(d-a)(\overline{b-a})+2(\overline{d-a})(b-a)+4(b-a)(\overline{b-a})=0 \\
& \Longleftrightarrow 2(d-a)(a-b)+2(\overline{d-a})(b-a) a b+4(b-a)(a-b)=0 \\
& \Longleftrightarrow(d-a)-(\overline{d-a}) a b+2(b-a)=0 \\
& \Longleftrightarrow 2 c-b-a-2 \bar{c} a b+a+b+2(b-a)=0 \\
& \Longleftrightarrow c^{2}-a b+b c-a c=0 \\
& \Longleftrightarrow(b+c)(c-a)=0,
\end{aligned}
$$

implying $c=-b$ and that triangle $A B C$ is right-angled at $A$.
We use areal co-ordinates with reference to the triangle $A B C$. Recall that if $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ are points in the plane, then the square of the distance between these two points is $-a^{2} v w-b^{2} w u-c^{2} u v$, where $(u, v, w)=\left(x_{1}-x_{2}, y_{1}-y_{2}, z_{1}-z_{2}\right)$.

In our case $A=(1,0,0), B=(0,1,0), C=(0,0,1)$, so $E=(3,0,2)$ and, introducing point $F$ as in the first solution, $F=(-1,0,2)$. Then

$$
B E^{2}=A D^{2} \Longleftrightarrow-2 a^{2}+6 b^{2}+3 c^{2}=2 a^{2}+2 b^{2}-c^{2},
$$

and thus $a^{2}=b^{2}+c^{2}$. Therefore $\angle B A C$ is a right angle by the converse of the theorem of Pythagoras.