# 2007 Indonesia MO Problems/Problem 7

## Problem

Points $A,B,C,D$ are on circle $S$, such that $AB$ is the diameter of $S$, but $CD$ is not the diameter. Given also that $C$ and $D$ are on different sides of $AB$. The tangents of $S$ at $C$ and $D$ intersect at $P$. Points $Q$ and $R$ are the intersections of line $AC$ with line $BD$ and line $AD$ with line $BC$, respectively.

(a) Prove that $P$, $Q$, and $R$ are collinear.

(b) Prove that $QR$ is perpendicular to line $AB$.

## Solution $[asy] size(225); pair a=(-50,0),b=(50,0),c=(-48,14),d=(-30,-40),o=(0,0); draw(circle(o,50)); draw(a--c--b--d--a); dot(a); label("A",a,NE); dot(b); label("B",b,E); dot(c); label("C",c,NE); dot(d); label("D",d,S); dot(o); label("O",o,SE); pair r=(-57.692,15.385),q=(-57.692,-53.846); draw (r--b--q); dot(r); label("R",r,NW); dot(q); label("Q",q,SW); draw(c--q); draw(r--d); draw(r--q); draw(a--b,dotted); draw(c--o--d--c,dotted); [/asy]$ Let $O$ be the center of the circle. Let $\angle CBA = b$ and $\angle DBA = a$. To prove that $P$ is on line $QR$, we can show that $\angle PRD = \angle QRD$.

Because $AB$ is a diameter, we know that $\angle ACB = 90^\circ$ and $\angle ADB = 90^\circ$. By the Vertical Angle Theorem, $\angle RAC = \angle QAD$, so by AA Similarity, $\triangle RAC \sim \triangle QAD$. Thus, $\frac{RA}{AC} = \frac{QA}{AD}$. Because $\angle RAQ = \angle CAD$, by SAS Similarity, $\triangle RAQ \sim \triangle CAD$, so $\angle ARQ = \angle ACD$. Since $\angle ACD$ and $\angle ABD$ are both inscribed angles with the same arc of a given circle, $\angle ABD = \angle ACD = a$, so $\angle ARQ = a$. $[asy] size(225); pair a=(-50,0),b=(50,0),c=(-48,14),d=(-30,-40),o=(0,0); draw(circle(o,50)); draw(a--c--b--d--a); dot(a); label("A",a,NE); dot(b); label("B",b,E); dot(c); label("C",c,NE); dot(d); label("D",d,S); dot(o); label("O",o,SE); pair r=(-57.692,15.385),q=(-57.692,-53.846),p=(-57.692,-19.231); dot(r); label("R",r,NW); dot(q); label("Q",q,SW); dot(p); label("P",p,W); draw(c--p--d); draw(r--b--q); draw(c--q); draw(r--d); draw(r--p); draw(a--b,dotted); draw(c--o--d--c,dotted); [/asy]$

Because $PC$ and $PD$ are both tangents to the circle, we must have $\angle ODP = \angle OCP$. Therefore, $\angle RDP = \angle BDO = a$ and $\angle PCQ = \angle OCB = b$. Additionally, from a property of inscribed angles, we must have $\angle ACD = \angle ABD = a$ and $\angle ADC = \angle ABC = b$. Thus, since the sum of the angles in a triangle is $180^\circ$, $\angle CPD = 180-(2a+2b)$.

Additionally, since $\triangle RDB$ is a right triangle, we must have $\angle BRD = 90-(a+b)$. Because $\angle CRD = \tfrac12 \cdot \angle CPD$ and $CP = PD$, we know that $C,R,D$ are in a circle with center $P$, so $RP = PC = PD$. Since $RB$ is a line, $\angle RCP = 90-b$, so by the Base Angle Theorem, $\angle CRP = \angle RCP = 90-b$. Thus, from the Angle Addition Postulate, $\angle PRD = (90-b)-(90-a-b) = a$.

Thus, we proved that $P$ is on line $RQ$. Additionally, by letting $K$ be the intersection of $RP$ and $AB$, we must have $\angle KRB = 90-b$ and $\angle RBK = b$, so $\angle RKB = 90^\circ$. By definition, $AB \perp RP$, and since $P$ is on line $RQ$, $AB \perp RQ$.