IRL ABooklet

Note that $|CD| = |CE|$ and $|RN| = |RE|$. Hence, triangles $\triangle ENR$ and $\triangle EDC$ which share an angle at $E$, are both isosceles and hence similar. In particular, $RN \parallel CB$ and
$$\angle CDE = \angle CED = \angle ENR = 90^\circ - \frac{1}{2}\angle C.$$

a.

Solution 1. Let $N'$ be the intersection of $BI$ and $DE$. Then
$$\begin{aligned} 180^\circ - \angle IN'E &= \angle BN'D = \angle CDE - \angle CBI = 90^\circ - \frac{1}{2}\angle C - \frac{1}{2}\angle B \\ &= \frac{1}{2}\angle A = \angle IAE \end{aligned}$$
hence $AIN'E$ is cyclic and we have $\angle IN'A = \angle IEA = 90^\circ$.

Let $AN'$ extended meet $BC$ at $K'$. Then, in $\triangle BAK'$, $BN'$ is both, angle bisector and altitude, which implies that $\triangle BAK'$ is isosceles and $N'$ is the midpoint of $AK'$. Hence $RN' \parallel CK'$ (midline) and so $\triangle EN'R \sim \triangle EDC$. Because $|CD| = |CE|$ we therefore also have $|RE| = |RN'|$, hence $N' = N$.

Solution 2. Let $K$ be the intersection of $AN$ with $BC$. Then $RN$ is the midline of $\triangle AKC$, so $N$ is the midpoint of $AK$, and $|RE| = |RN| = \frac{1}{2}|CK|$. Since
$$a = |BC| = |BD| + |CE|, \ b = |CA| = |AE| + |CE|, \ c = |AB| = |AE| + |BD|$$

$$|AE| = \frac{b+c-a}{2} \quad \text{and}$$
$$|RE| = |RA| - |AE| = \frac{b}{2} - \frac{b+c-a}{2} = \frac{a-c}{2}.$$

Hence, $|CK| = a - c$ and so $|BK| = c$. This shows that $\triangle BAK$ is isosceles and $BN$ is a median of it, hence $BN$ is also an angle bisector and thus passes through the incentre $I$, and $BN$ is an altitude so that $\angle BNA = 90^{\circ}$.

b.

Solution 1. Extend $RQ$ to meet $BC$ at $M$. Since $NR \parallel MD$ and $|NQ| = |QD|$, we see that $\triangle QNR \equiv \triangle QDM$. This implies that $Q$ is the midpoint of $RM$ and $|DM| = |RN| = |RE|$.

Because we also have $|ID| = |IE|$ and $\angle IDM = \angle IER = 90^{\circ}$, we now get $\triangle IDM \equiv \triangle IER$, hence $|IR| = |IM|$. This implies that the median $IQ$ in the isosceles triangle $\triangle IRM$ is also a perpendicular bisector, i.e. $IQ \perp RM$. Note that this proof, which is independent of (a), does not require $R$ to be the midpoint of $AC$.

Solution 2. Note that $\triangle IDN$ is similar to $\triangle ICA$ since
$$\angle IND = \frac{1}{2}\angle A = \angle IAC \quad \text{and}$$
$$\angle IDN = 90^{\circ} - \angle EDC = \frac{1}{2}\angle C = \angle ICA.$$
Since $Q$ is the midpoint of $DN$ and $R$ the midpoint of $CA$ it follows that $\triangle IQN \sim \triangle IRA$, in spiral symmetry and hence also $\triangle IQR \sim \triangle INA$. Using part (a), it now follows that $\angle IQR = \angle INA = 90^{\circ}$.

Solution 3. Let $S$ on $IQ$ extended be such that $Q$ is the midpoint of $IS$. Then $IDSN$ is a parallelogram (diagonals bisect each other) and $|SN| = |DI|$, as well as $SN \parallel DI \perp BC \parallel NR$.

By SAS we now obtain $\triangle SNR \equiv \triangle IER$, thus $|RS| = |RI|$. In isosceles triangle $\triangle IRS$, $RQ$ is a median, hence also a perpendicular bisector, thus $\angle RQI = 90^\circ$.