IRL_ABooklet_2023

a. We denote the angles of $△ABC$ by $\angle A$, $\angle B$, $\angle C$. Since $AI$ is a diameter, we have $\angle ACI=\angle ABI=90^{\circ }$. Also, $\angle AIC=\angle B$ (both standing on arc $AC$), and $\angle CAI=\angle CBI$ (both standing on arc $CI$). Therefore,



$△ACI\sim △ADB\sim △BQI$. Similarly, $△ABI\sim △ADC\sim △CQI$ and $$\frac{|AC|}{|CI|}=\frac{|AD|}{|DB|}=\frac{|BQ|}{|QI|}\text{as well as}\frac{|AB|}{|BI|}=\frac{|AD|}{|DC|}=\frac{|CQ|}{|QI|}$$ which gives $|AD|\cdot |QI|=|CD|\cdot |CQ|=|BD|\cdot |BQ|$.

b. Solution 1: Let $X$ denote the intersection point of $AI$ and $BC$. Then, by power of $X$ with respect to the circumcircle of $ABC$ (or $△AXB\sim △CXI$) we have $|AX|\cdot |XI|=|CX|\cdot |XB|$. By power of $X$ with respect to the circumcircle of $CBO$, we have $|OX|\cdot |XP|=|CX|\cdot |XB|$. Combining both equations, we obtain $$|OX|\cdot |XP|=|AX|\cdot |XI|\text{i.e.}\frac{|AX|}{|OX|}=\frac{|XP|}{|XI|}$$



By the Intercept Theorem, or $△OMX\sim △ADX$, we have $$\frac{|AX|}{|OX|}=\frac{|DX|}{|MX|},\text{hence}\frac{|XP|}{|XI|}=\frac{|DX|}{|MX|}$$ This implies similarity of triangles $MXI$ and $DXP$, hence $MI\parallel DP$.

Solution 2: Substituting $|CD|=|BC|-|BD|$ and $|BQ|=|BC|-|CQ|$ in the equation $|CD|\cdot |CQ|=|BD|\cdot |BQ|$ from part (a) yields $|BD|=|CQ|$. Since $M$ is the midpoint of $BC$, it is then also the midpoint of $DQ$. We define $J$ on the extended line $QI$ such that $I$ is the midpoint of segment $QJ$. Then $MI\parallel DJ$ (mid-line in $△DQJ$). It remains to prove that $J$ is on the line $DP$.



We first prove that $I$ is the incentre of $△CPB$. Indeed, we have $\angle CBP=\angle COP$ and $\angle BCP=\angle BOP$ in $\omega$ $\angle COP=2\angle CBI$ and $\angle BOP=2\angle BCI$ in $\Gamma$ so $\angle CBP=2\angle CBI$ and $\angle BCP=2\angle BCI$ hence $BI$ and $CI$ are angle bisectors and $I$ is the incentre of $△CPB$. Because $AI$ is a diameter of $\Gamma$, $AC\perp CI$ and $AB\perp BI$. Therefore, $AC$ and $AB$ are external angle bisectors of $△CPB$ and so $A$ is an excentre.

Consider the inradius and exradius given by $IN$ and $AL$, both perpendicular to $PC$. Then triangles $△PNI$ and $△PLA$ are similar, which implies $$\frac{|PI|}{|PA|}=\frac{|IN|}{|AL|}=\frac{|IQ|}{|AD|}=\frac{|IJ|}{|AD|}.$$ Because $IJ$ is parallel to $AD$, we have $\angle PIJ=\angle PAD$. Hence, triangles $△PIJ$ and $△PAD$ are similar and $P$, $J$, $D$ are collinear.

Solution 3: From (a) we get $|BD|=|CQ|$, see Solution 2. This implies $|BQ|=|CD|$ and that $M$ is the midpoint of $QD$. We extend the line $QI$ to intersect $PD$ at $J$. To prove $MI\parallel PD$, it is sufficient to show $|QI|=|IJ|$. This is equivalent to $\frac{|QI|}{|AD|}=\frac{|PI|}{|PA|}$, because $\frac{|IJ|}{|AD|}=\frac{|PI|}{|PA|}$ follows from $QJ\parallel AD$.



Using (a), we obtain $$\frac{|QI|}{|AD|}=\frac{|BD|\cdot |BQ|}{|AD{|}^2}=\frac{|BD|\cdot |CD|}{|AD{|}^2}=\cot B\cot C.$$

On the other hand, $$\frac{|PI|}{|PA|}=\frac{[CPI]}{[CPA]}=\frac{\frac{1}{2}|CP|\cdot |CI|\sin \angle PCI}{\frac{1}{2}|CP|\cdot |CA|\sin \angle PCA}=\frac{|CI|\sin \angle PCI}{|CA|\sin \angle PCA},$$ and $\frac{|CI|}{|CA|}=\cot \angle CIA=\cot B$, while $$\begin{array}{rl}\angle PCA& =180^{\circ }-\angle CAO-\angle CPO=180^{\circ }-\angle ACO-\angle CBO\\ & =180^{\circ }-\angle ACO-\angle OCB=180^{\circ }-\angle C\end{array}$$ and so $\angle PCI=\angle PCA-90^{\circ }=90^{\circ }-\angle C$. Hence $$\frac{\sin \angle PCI}{\sin \angle PCA}=\frac{\sin (90^{\circ }-\angle C)}{\sin (180^{\circ }-\angle C)}=\frac{\cos C}{\sin C}=\cot C.$$ Therefore, $$\frac{|PI|}{|PA|}=\frac{|CI|}{|CA|}\cdot \frac{\sin \angle PCI}{\sin \angle PCA}=\cot B\cot C=\frac{|QI|}{|AD|},$$