Euclidean geometry of the ratio of two lengths

Question

Let ABC be a triangle with $AB=BC$ and $\angle{ABC}=90^{\circ}$. Let $D$ be the midpoint of $AC$, and $E$ be a point on the opposite of $AC$ as $B$ and $\angle{AEC}=45^{\circ}$.

Is the ratio of $EB/ED$ constant? If so, what is its value?

Answer 1

Method 1. We will use the method of coordinates. Place the origin at point $D$ so that the $x$-axis coincides with line $DC$ (direction from point $D$ to point $C$) and $y$-axis coincides with line $DB$ (direction from point $D$ to point $B$). Then $D=(0,0)$, $B=(0,a)$, $C=(a,0)$, $A=(-a,0)$, $a>0$. Point $E$ belongs to a circle with center at point $O=(0,-a)$ (because $O$ is middle point of $CE_1$, where $E_1$ is fixed point such that $CE_1=CA$, $CE_1 \bot CA$ and $y_{E_1}<0$) and radius equal to $\sqrt{2}a$, so: $$ x_E^2+(y_E+a)^2=2a^2, \quad y_E<0.$$ Now everything is ready to find the ratio $$\frac{EB^2}{ED^2}=\frac{x_E^2+(y_E-a)^2}{x_E^2+y_E^2}=\frac{2a^2-(y_E+a)^2+(y_E-a)^2}{2a^2-(y_E+a)^2+y_E^2}=$$ $$=\frac{2a^2-4y_Ea}{a^2-2y_Ea}=2,$$ because $a>0$ and $y_E \neq \frac{a}{2}$.

Method 2. We will say that points $A$ and $B$ are symmetric with respect to the circle $\omega$ if points $A$ and $B$ pass into each other upon inversion with respect to the circle $\omega$.

LEMMA 1. Let the points $A$ and $B$ be symmetric with respect to the circle $\omega$. Then for any point $M$ of the circle $\omega$ the ratio $MA:MB$ will be constant.

PROOF. Let $P$ and $Q$ be the points of intersection circle $\omega$ and line $AB$. Since $A$ and $B$ are symmetric $OB\cdot OA=R^2$. Then $$AB\cdot OB =(OA-OB)\cdot OB=OB\cdot OA-OB^2= R^2-OB^2=(R-OB)\cdot (R+OB)=BP\cdot BQ=BL\cdot BM.$$ So, $\frac{AB}{BL}=\frac{BM}{BO}$ and therefore triangles $AMB$ and $LOB$ are similar and $\angle AMB=\angle LOB$. Hence, $\angle LMK=\angle LOQ$, arc $LQ$ is equal to arc $KQ$ and $MQ$ is bissector of angle $LMK$. Because $\angle PMQ= 90^{\circ}$, $MP$ is bissector of angle $AMB$. Finally we get that $\omega$ is Apollonius circle. This follows from a purely geometric approach to the construction of Apollonius circle (see http://jwilson.coe.uga.edu/emt725/Apollonius/Cir.html). By definition (see https://en.m.wikipedia.org/wiki/Circles_of_Apollonius) it follows that $MA:MB$ is constant.

END OF PROOF.

Now return to our problem. We will use first picture. Note that $\angle BAO=\angle BCO= 90^{\circ}$ and hence $B$ and $D$ are symmetrical points with respect to circle $\omega$. This follows from the method of constructing the image of a point in inversion with respect to a circle (see https://en.m.wikipedia.org/wiki/Inversive_geometry).

Hence, from LEMMA 1 it follows that $\frac{EB}{ED}$ is constant for any position of point $E$ on the circle, including when point $E$ coincides with point $C$. For this trivial case we can easily find the ratio $\frac{CB}{CD}=\sqrt{2}$.

Answer 2

Assume $ AD=DC=1$ with D as origin.

Straightaway mark coordinates of the five points and find distances by the distance formula.

$$\frac{EB}{ED}=\frac{\sqrt 10}{\sqrt5}=\sqrt 2. ~$$