# 2002 Pan African MO Problems/Problem 2

## Problem $\triangle{AOB}$ is a right triangle with $\angle{AOB}=90^{o}$. $C$ and $D$ are moving on $AO$ and $BO$ respectively such that $AC=BD$. Show that there is a fixed point $P$ through which the perpendicular bisector of $CD$ always passes.

## Solution

We can consider two cases: one where the legs have equal length and one where the legs don’t have equal length.

Case 1: $AO = OB$

Since $AO = OB$ and $AC = BD$, by the Segment Addition Postulate, $OC = OD$. This, $\triangle OCD$ is a 45-45-90 triangle.

Let $E$ be the midpoint of $CD$. By SSS Congruency, $\triangle OCE \cong \triangle ODE$. Thus, $\angle OEC = \angle OED$, so $OE \perp CD$.

Now extend $OE$ to $AB$, and let $F$ be the point of intersection. By SAS Similarity, $\triangle AOB \sim \triangle COD$, so $\angle BAO = \angle DCO$ and $AB \parallel CD$. Thus, $OF \perp AB$. $[asy] pair a=(0,10),o=(0,0),b=(10,0),c=(0,8),d=(8,0),e=(4,4); draw(a--o--b--a); dot(a); label("A",a,NW); dot(o); label("O",o,SW); dot(b); label("B",b,SE); dot(c); label("C",c,W); dot(d); label("D",d,S); draw(c--d,dotted); draw(o--(5,5),dotted); dot((4,4)); label("E",(4,4),W); dot((5,5)); label("F",(5,5),NE); [/asy]$ By HL Congruency, $\triangle AOF \cong \triangle BOF$, so $AF = BF$. Therefore, $OF$ is a perpendicular bissector of $AB$. Thus, all perpendicular bissectors of $CD$ where $AO = OB$ are also a perpendicular bisector of $AB$, so there is a point $P$ where the perpendicular bissector of $CD$ passes through.

Case 2: $AO \ne OB$

WLOG, let $AO > OB$. Additionally, let $A = (0,2a)$ and $B = (2b,0)$, so $a > b$. $[asy] pair a=(0,20),o=(0,0),b=(10,0); draw(a--o--b--a); dot(a); label("A",a,NW); dot(o); label("O",o,SW); dot(b); label("B",b,SE); [/asy]$ Let $D_1, C_1$ be points where $D_1 = O$, making $BD_1 = 2b$. Thus, $AC_1 = 2a-2b$, so $C_1 D_1$ is on $(0,2a-2b)$. Since the midpoint of $C_1 D_1$ is on $(0,a-b)$, the y-coordinate of $P$ must equal $a-b$.

Let $D_2, C_2$ be points where $D_2 = B$, making $AC_2 = BD_2 = 0$. The midpoint of $C_2 D_2$ is $(b,a)$, and the slope of $C_2 D_2$ is $- \tfrac{a}{b}$. Thus, the equation of the line perpendicular to $C_2 D_2$ is $y - a = \tfrac{b}{a} (x - b)$. Since the y-coordinate of $P$ equals $a-b$, substituting and solving for $x$ results in $x = b-a$. The coordinates of $P$ are $(b-a,a-b)$; now we need to prove that this point is on any perpendicular bissector of $CD$.

Let $D_3, C_3$ be any point of $D, C$ at a given time, and let the coordinates of $D_3$ be $(2b-2n,0)$. That means the coordinates of $C_3$ are $(0,2a-2n)$. The midpoint of $C_3 D_3$ is $(b-n,a-n)$, and the slope of $C_3 D_3$ is $\tfrac{n-a}{b-n}$. Thus, the equation of the line perpendicular to $C_3 D_3$ is $y - a+n = \tfrac{n-b}{n-a} (x-b+n)$. Plugging in $b-a$ for $x$ means that $y = a-b$, so point $P$ is on the line.

Thus, there is a fixed point $P$ through which the perpendicular bisector of $CD$ always passes.